Chapter 13: AC Circuits
In the late 1880s, the United States was divided over a simple question: what kind of electricity should light up its cities? Thomas Edison had built his early power stations around direct current, in which the polarity of the supply never changes. His rival system, backed by George Westinghouse and developed largely from the ideas of Nikola Tesla, used alternating current, in which the voltage reverses direction again and again, many times every second.
The disagreement became so bitter that it is remembered as the “War of the Currents.” Edison’s argument was about safety. Tesla and Westinghouse had physics on their side: alternating voltage can be stepped up to very high values for transmission and stepped down again near homes, so the same power can be sent over long distances through thinner wires with far smaller heating losses. Direct current at that time could not be transformed so easily.
The dispute was settled by a river. In 1895, the hydroelectric power station at Niagara Falls began generating alternating current, and soon it was carrying electrical energy over about 40 kilometres to the city of Buffalo. Long-distance AC transmission had been proved to work, and the world’s power grids followed.
Every socket in your home is a descendant of that decision. But a supply whose voltage keeps reversing behaves in strange ways once you connect coils and capacitors to it — and that is what this chapter is about.
Diagram
Fig. 13.0 – Opening illustration: the Niagara Falls generating station with transmission lines running to a distant city, and an inset sine-wave voltage curve reversing about the zero line.
By the end of this chapter you will be able to answer each of these questions using the physics of alternating currents.
13.1 Introduction
In school you have already learnt that electricity can be supplied to a circuit in two different ways. The difference between them lies in one thing only: whether the polarity of the supply stays fixed or keeps reversing.
Direct Current and Alternating Current
In a direct current (DC) supply, the polarity of the voltage is fixed. The positive and negative ends of the power supply do not change places, so the current always flows through the circuit in the same direction. A dry cell connected to a bulb is the simplest example.
In an alternating current (AC) supply, the polarity of the voltage keeps changing periodically. The terminal that is positive at one instant becomes negative a fraction of a second later, and then positive again. The current therefore flows first one way through the circuit and then the other, over and over again. Notice in Fig. 13.1a how a DC voltage stays on one side of the zero line while an AC voltage crosses it repeatedly.
Diagram
Fig. 13.1a – Two graphs of voltage versus time side by side: a flat horizontal line above the axis for DC, and a sinusoidal curve crossing the time axis for AC.
Why Our Homes Are Supplied with AC
You have studied the generation of AC voltage in the previous chapter. Almost all the electricity that reaches homes, schools and factories is supplied as AC, and there are two practical reasons for this: it is cheap to generate, and it is convenient to transport over long distances.
The convenience comes from the fact that an alternating voltage can be raised or lowered easily using a transformer, which works only when the current through it is changing. A steady direct current produces no change of magnetic flux, so it cannot be stepped up or down in this simple way.
Devices That Use AC and Devices That Convert It
Some of the appliances we use at home or in offices — TV, computer, transistor, radio, and so on — cannot work on alternating voltage directly. They first convert AC into DC using a device called a rectifier, which you will study in Chapter 16, and only then use it.
However, there are many domestic devices such as a fan, fridge, air conditioner, induction heater and coil heater which run directly on AC, without any such conversion. Almost all these devices use components like an inductor and a capacitor in their circuits.
That is why the behaviour of these components on an alternating supply matters so much. In this chapter we will study the passage of AC through resistors, inductors and capacitors — first each of them separately, and then all of them together in one circuit.
13.2 AC Generator
In the last chapter we studied that the source of AC — the generator — produces a time dependent emf. “Time dependent” simply means that the value of the emf is not fixed: it changes from instant to instant. For an AC generator this emf is given by
\[e = e_0 \sin \omega t \qquad \text{--- (13.1)}\]
Here \(e\) is the instantaneous emf (its value at that particular instant), measured in volt (V). The symbol \(e_0\) is the peak value of the emf, that is, the largest value it reaches, also in volt. The symbol \(\omega\) is the angular frequency of rotation of the coil in the AC generator, measured in radian per second (rad s\(^{-1}\)), and \(t\) is the time in second (s).
Where the Sine Form Comes From
The sine in Eq. (13.1) is not put there by hand — it follows from the way the generator is built. Inside the generator a coil of \(N\) turns and area \(A\) is rotated steadily in a uniform magnetic field \(B\), as shown in Fig. 13.1b.
Diagram
Fig. 13.1b – A rectangular coil rotating between the poles of a magnet, with slip rings and brushes taking the output to an external circuit.
As the coil turns, the angle between its normal and the field keeps changing, so the magnetic flux linked with the coil keeps changing as \(\phi = NBA\cos\omega t\). By Faraday’s law the induced emf equals the negative rate of change of this flux, and differentiating gives \(e = NBA\,\omega \sin \omega t\). Comparing this with Eq. (13.1), the peak value is \(e_0 = NBA\omega\). This is why a faster rotation gives both a larger peak emf and a higher frequency.
Current Produced by This emf
Since the current in the circuit is driven by this emf, its time variation is similar to that of the emf. The current in a circuit connected to such a generator is therefore of the form
\[i = i_0 \sin (\omega t + \alpha)\]
where \(i_0\) is the peak value of the current, measured in ampere (A), and \(\alpha\) represents the phase difference between the current \(i\) and the emf \(e\), measured in radian.
The extra angle \(\alpha\) deserves attention. It says that the current need not reach its maximum at the same instant as the emf — it may run ahead of the emf or lag behind it. Everything you study in the rest of this chapter depends on what value \(\alpha\) takes for a given circuit element.
Why This emf Is Called Alternating
From Eq. (13.1) it can be seen that the induced emf varies sinusoidally with time, as shown in the graph of \(e\) versus \(\omega t\) in Fig. 13.1.
Diagram
Fig. 13.1 – Graph of e versus ωt: a sine curve rising to \(e_0\) at \(\pi/2\), crossing zero at \(\pi\), reaching its minimum at \(3\pi/2\) and returning to zero at \(2\pi\).
From the graph it can be seen that the emf is positive for the first half of the cycle and negative for the next half. In other words, the direction of the emf is reversed after every half revolution of the coil. This type of emf is called alternating emf, and the corresponding current is called alternating current.
In numerical problems, a question usually gives you an equation such as \(e = 6\sin(100\pi t)\) and expects you to read off the physical quantities by comparing it term by term with Eq. (13.1). The number multiplying the sine is \(e_0\), and the coefficient of \(t\) inside the bracket is \(\omega\), from which the frequency follows using \(\omega = 2\pi f\). This comparison step is worth practising until it is automatic.
13.3 Average and RMS Values
An alternating voltage or current does not stay at one value. In a single cycle it passes through every value between zero and the peak value, in one direction, and then through the same set of values in the opposite direction.
This creates a practical difficulty. If someone asks “how large is this current?”, there is no single obvious number to give, because the answer changes from instant to instant. So we define a few specific values which are convenient for describing and comparing voltage or current waveforms. The first of these is the peak value.
The peak value is written as \(i_0\) for current (in ampere, A) and \(e_0\) for emf (in volt, V). It is a useful number, but by itself it does not tell us how much heating or how much work the current can actually deliver, because the current stays at its peak only for an instant. That is why two more values are needed: the average value and the rms value.
Average Value of AC
The average value of AC is the average of all the values of the voltage (or current) taken over one half cycle.
Why only half a cycle? Look again at the graph of \(e\) versus \(\omega t\) in Fig. 13.1. For the first half of the cycle the curve lies above the axis and for the second half it lies an equal amount below it. The positive and negative halves cancel exactly, so the average value of \(\sin \omega t\) over a full cycle is zero. Hence the average over a full cycle is always zero, and the mean value of AC over a complete cycle has no significance. For this reason the mean value of AC is defined as the average over a half cycle only, as shown by the shaded region in Fig. 13.1c.
Diagram
Fig. 13.1c – One full sine cycle with the first half cycle shaded, marked “average taken over this half only”; the second half marked “cancels the first over a full cycle”.
To find this average we take the average value of \(\sin\theta\) for \(\theta\) in the range \(0\) to \(\pi\). Averaging a continuous quantity means dividing the area under its curve by the length of the interval, which is exactly what the following ratio of integrals does:
\[\text{Average of } \sin\theta = \frac{\int_0^{\pi} \sin\theta \; d\theta}{\int_0^{\pi} d\theta} = \frac{\left[-\cos\theta\right]_0^{\pi}}{\left[\theta\right]_0^{\pi}} = \frac{2}{\pi} = 0.637\]
Therefore the average value of an alternating current or emf is \(0.637\) times its peak value:
\[i_{av} = 0.637\, i_0 \qquad \text{and} \qquad e_{av} = 0.637\, e_0\]
where \(i_{av}\) and \(e_{av}\) are the average values of the alternating current (in ampere) and of the alternating emf or voltage (in volt) respectively.
RMS Value
A moving coil ammeter and a moving coil voltmeter measure the average value of the current through them and of the voltage applied across them respectively. Since the average of an alternating quantity over a full cycle is zero, such an instrument would simply show no deflection. It is obvious, therefore, that moving coil instruments cannot be used to measure alternating current and voltage.
To measure these quantities we must make use of some property of the current that does not depend on the changes in its direction. The heating effect is exactly such a property. Heat produced depends on the square of the current, and the square of a number is always positive whether the number is positive or negative. So the heating does not depend on the direction of flow of the current.
Diagram
Fig. 13.1d – Two curves on the same time axis: \(i\) versus \(\omega t\) swinging above and below zero, and \(i^2\) versus \(\omega t\) lying entirely above the axis with its mean value marked at \(i_0^2/2\).
This suggests a fair way of stating “how big” an alternating current is: compare it with the steady current that would heat the same resistor equally. Consider an alternating current of peak value \(i_0\) flowing through a resistance \(R\), and let \(H\) be the heat produced in time \(t\). Now the same quantity of heat \(H\) can be produced in the same resistance \(R\) in the same time \(t\) by passing a steady current of constant magnitude through it. The value of such a steady current is called the effective value, or virtual value, or rms value of the given alternating current, and is denoted by \(i_{rms}\).
The name describes the recipe in reverse order: square the current at every instant, take the mean of those squares over a full cycle, and then take the square root. Carrying this out for \(i = i_0 \sin\omega t\) gives
\[i_{rms} = \sqrt{\frac{\int_0^{2\pi} i_0^{2} \sin^{2}\omega t \; d(\omega t)}{\int_0^{2\pi} d(\omega t)}} = \sqrt{\frac{i_0^{2}}{2\pi} \int_0^{2\pi} \frac{(1 - \cos 2\omega t)}{2} \; d(\omega t)} = \sqrt{\frac{i_0^{2}}{2}}\]
The step in the middle uses the identity \(\sin^{2}\omega t = \dfrac{1 - \cos 2\omega t}{2}\). The \(\cos 2\omega t\) part averages to zero over a full cycle, leaving only the constant \(\tfrac{1}{2}\), which is the same as saying that the average value of \(\sin^{2}\omega t\) over a cycle is \(\tfrac{1}{2}\). Hence
\[i_{rms} = \frac{i_0}{\sqrt{2}} = 0.707\, i_0\]
Similarly it can be shown that for the alternating emf
\[e_{rms} = \frac{e_0}{\sqrt{2}} = 0.707\, e_0\]
Heat Produced Over a Complete Cycle
The result above can be checked directly by calculating the heat produced by a sinusoidally varying AC over a complete cycle. Using \(H = \int i^{2} R \, dt\) with \(i = i_0 \sin \omega t\) and integrating over one full period, during which \(\omega t\) goes from \(0\) to \(2\pi\),
\[H = \int_0^{2\pi/\omega} i_0^{2} \sin^{2}(\omega t)\, R \; dt = \frac{i_0^{2} R}{2} \times \frac{2\pi}{\omega}\]
\[\therefore H = i_{rms}^{2} R \, t \qquad \left(\because i_{rms} = \frac{i_0}{\sqrt{2}}\right)\]
It is the same as the heat produced by a DC current of magnitude \(i_{rms}\) flowing for time \(t = \dfrac{2\pi}{\omega}\). This is precisely what the definition of the rms value promised, so the two routes agree.
13.4 Phasors
The study of AC circuits becomes much simpler if we represent the alternating current and the alternating emf as rotating vectors, drawn with the angle between them equal to the phase difference between the current and the emf. These rotating vectors are called phasors.
Why a Rotating Vector Helps
The reason for this device is practical. In an AC circuit we constantly need to add voltages that are out of step with one another — for example the voltage across a coil and the voltage across a capacitor in the same circuit. Two sine functions with different phases cannot simply be added like ordinary numbers, because their peaks occur at different instants.
A phasor turns that problem into a geometry problem. Once each sinusoidal quantity is drawn as an arrow, the arrows can be combined by the ordinary rules of vector addition, which you already know.
How a Phasor Is Drawn
The phasors for the alternating emf and the alternating current are inclined to the horizontal axis at an angle \(\omega t\) or \(\omega t + \alpha\), and they rotate in the anticlockwise direction. The angle grows steadily with time because \(\omega\) is the angular frequency in radian per second, so the phasor makes one complete revolution in exactly one cycle of the AC.
The length of the arrow represents the maximum value of the quantity, that is, \(i_0\) for the current and \(e_0\) for the emf. The length therefore stays fixed while the arrow turns.
The projection of the vector on a fixed axis gives the instantaneous value of the alternating current or the alternating emf. Which axis is used depends on how the quantity has been written. In the sine form, \(i = i_0 \sin \omega t\) or \(e = e_0 \sin \omega t\), the projection is taken on the Y-axis, as shown in Fig. 13.2 (a). In the cosine form, \(i = i_0 \cos \omega t\) or \(e = e_0 \cos \omega t\), the projection is taken on the X-axis, as shown in Fig. 13.2 (b).
Diagram
Fig. 13.2 (a) and (b) – Phasor diagrams: (a) an arrow of length \(e_0\) at angle \(\omega t\) with its projection dropped on the Y-axis; (b) the same arrow with its projection taken on the X-axis.
You can check that this works. When the arrow of length \(e_0\) stands at angle \(\omega t\), its vertical projection is \(e_0 \sin \omega t\), which is exactly the instantaneous emf. As the arrow sweeps round once, the projection rises to \(+e_0\), falls back through zero to \(-e_0\), and returns — tracing out one full sine wave, as shown in Fig. 13.2 (c).
Diagram
Fig. 13.2 (c) – A rotating phasor on the left and the sine curve it generates on the right, with dotted lines linking four positions of the arrow to the corresponding points on the curve.
Because the two phasors rotate together at the same angular frequency, the angle between them never changes. That fixed angle is the phase difference \(\alpha\), and it is the single most important piece of information in an AC circuit diagram.
The representation of harmonically varying quantities as rotating vectors is what enables us to use the laws of vector addition for adding these quantities. Every phasor diagram in the rest of this chapter rests on this one idea.
13.5 Different Types of AC Circuits
In this section we will derive the voltage–current relations for individual as well as combined circuit elements — resistors, inductors and capacitors — carrying a sinusoidal current. Unless otherwise specified, we assume the capacitor and the inductor to be ideal, which means the inductor has no resistance of its own and the capacitor has no leakage.
The plan is to take each element separately first, find out how much it opposes the current and by what phase angle it shifts the current with respect to the applied emf, and only then put all three together in one circuit.
AC Voltage Applied to a Resistor
Suppose a resistor of resistance \(R\) is connected to an AC source of emf whose instantaneous value \(e\) is given by
\[e = e_0 \sin \omega t \qquad \text{--- (13.2)}\]
where \(e_0\) is the peak value of the voltage in volt (V) and \(\omega\) is its angular frequency in rad s\(^{-1}\). The arrangement is shown in Fig. 13.3.
Diagram
Fig. 13.3 – An AC source, drawn as a circle with a sine symbol inside, connected in a single loop to a resistor R.
Let \(e\) also be the potential drop across the resistance. Then, by Ohm’s law,
\[e = iR \qquad \text{--- (13.3)}\]
since the instantaneous emf of the source must equal the instantaneous value of the potential drop across the resistor. There is nothing else in the loop to share the voltage.
From Eq. (13.2) and Eq. (13.3) we have
\[iR = e = e_0 \sin \omega t\]
\[\therefore \; i = \frac{e_0 \sin \omega t}{R}\]
\[\therefore \; i = i_0 \sin \omega t \qquad \left( \because i_0 = \frac{e_0}{R} \right) \qquad \text{--- (13.4)}\]
Comparing \(i_0 = \dfrac{e_0}{R}\) with Ohm’s law, we find that resistors behave in the same way for both AC and DC voltage. Hence the behaviour of \(R\) in DC and in AC circuits is the same, and \(R\) can reduce DC as well as AC equally effectively.
From Eq. (13.2) and Eq. (13.4) we see that for a resistor there is zero phase difference between the instantaneous alternating current and the instantaneous alternating emf — that is, they are in phase. Both \(e\) and \(i\) reach their zero, minimum and maximum values at the same time, as shown in Fig. 13.4.
Diagram
Fig. 13.4 – Graph of e and i versus ωt: two sine curves of different heights, crossing zero and peaking at exactly the same values of ωt, one drawn solid for e and one dashed for i.
Phasor Diagram for a Purely Resistive Circuit
In an AC circuit containing \(R\) only, the current and the voltage are in the same phase. Hence both phasors, the one for \(i\) and the one for \(e\), point in the same direction, making an angle \(\omega t\) with OX. Their projections on the vertical axis give their instantaneous values. The phase angle between the alternating current and the alternating voltage through \(R\) is zero, as shown in Fig. 13.5.
Diagram
Fig. 13.5 – Phasor diagram for a purely resistive load: two arrows of lengths \(e_0\) and \(i_0\) lying along the same line at angle ωt to the X-axis, with their projections marked on the Y-axis.
AC Voltage Applied to an Inductor
The activity above is worth thinking about before you read on: on a DC supply, a coil behaves no differently from a straight wire. We shall now see that on an AC supply it behaves very differently indeed.
Let us now connect the source of alternating emf to a circuit containing a pure inductor \(L\) only, as shown in Fig. 13.6. We assume that the inductor has negligible resistance, so the circuit is a purely inductive circuit. Suppose the alternating emf supplied is represented by
\[e = e_0 \sin \omega t \qquad \text{--- (13.5)}\]
Diagram
Fig. 13.6 – An AC source connected through a key k to a coil marked L, with the current direction i marked on both sides of the loop.
When the key k is closed, current \(i\) begins to grow in the inductor. Because the magnetic flux linked with the coil changes, an induced emf is produced which opposes the applied emf, in accordance with Faraday’s law.
According to Lenz’s law, this induced emf is
\[e' = -L \frac{di}{dt} \qquad \text{--- (13.6)}\]
where \(e'\) is the induced emf in volt, \(L\) is the self inductance in henry (H) and \(\dfrac{di}{dt}\) is the rate of change of current in ampere per second. The minus sign is the mathematical statement of Lenz’s law: the induced emf always acts against the change that is producing it.
To maintain the flow of current in the circuit, the applied emf \(e\) must be equal and opposite to the induced emf \(e'\). According to Kirchhoff’s voltage law, since there is no resistance in the circuit, the applied emf has nothing else to balance:
\[e = -e'\]
\[\therefore \; e = L \frac{di}{dt} \qquad \text{(from Eq. 13.6)}\]
\[\therefore \; di = \frac{e}{L}\, dt\]
Integrating the above equation on both sides, and substituting for \(e\) from Eq. (13.5), we get
\[\int di = \int \frac{e}{L}\, dt = \int \frac{e_0 \sin \omega t}{L}\, dt\]
\[\therefore \; i = -\frac{e_0}{\omega L}\cos \omega t + \text{constant}\]
The constant of integration is time independent and has the dimensions of current. As the emf oscillates about zero, the current \(i\) also oscillates about zero, so there cannot be any component of current which is time independent. Thus the integration constant is zero.
Using the trigonometric identity \(-\cos \omega t = \sin\left(\omega t - \dfrac{\pi}{2}\right)\), we can write
\[i = -\frac{e_0}{\omega L}\cos \omega t = \frac{e_0}{\omega L} \sin\left(\omega t - \frac{\pi}{2}\right)\]
\[\therefore \; i = i_0 \sin\left(\omega t - \frac{\pi}{2}\right) \qquad \text{--- (13.7)}\]
\[\text{where} \quad i_0 = \frac{e_0}{\omega L} \qquad \text{--- (13.8)}\]
Here \(i_0\) is the peak value of the current. Eq. (13.7) gives the alternating current developed in a purely inductive circuit when it is connected to a source of alternating emf.
Comparing Eq. (13.5) and Eq. (13.7), we find that the alternating current \(i\) lags behind the alternating emf \(e\) by a phase angle of \(\pi/2\) radian (\(90^\circ\)). The same fact can be stated the other way round: the voltage across \(L\) leads the current by a phase angle of \(\pi/2\) radian (\(90^\circ\)), as shown in Fig. 13.7.
Diagram
Fig. 13.7 – Graph of e and i versus ωt for a pure inductor: the current curve shifted a quarter cycle to the right of the emf curve, so that i is zero when e is maximum.
Phasor Diagram for a Purely Inductive Circuit
The phasor representing the peak emf \(e_0\) makes an angle \(\omega t\) in the anticlockwise direction from the horizontal axis. As the current lags behind the voltage by \(90^\circ\), the phasor representing \(i_0\) is turned clockwise through \(90^\circ\) from the direction of \(e_0\), as shown in Fig. 13.8.
Diagram
Fig. 13.8 – Phasor diagram for a purely inductive circuit: the phasor \(e_0\) at angle ωt and the phasor \(i_0\) drawn 90° clockwise from it.
Inductive Reactance
The opposing nature of an inductor to the flow of alternating current is called inductive reactance.
Comparing Eq. (13.8) with Ohm’s law written as \(i_0 = \dfrac{e_0}{R}\), we find that the quantity \(\omega L\) sits exactly where \(R\) used to sit. It therefore represents the effective resistance offered by the inductance \(L\). It is called the inductive reactance and is denoted by \(X_L\).
\[\therefore \; X_L = \omega L = 2\pi f L \qquad (\because \omega = 2\pi f)\]
where \(f\) is the frequency of the AC supply in hertz (Hz).
The function of the inductive reactance is similar to that of the resistance in a purely resistive circuit: it limits the current. It is directly proportional to the inductance \(L\) and to the frequency \(f\) of the alternating current.
The dimensions of inductive reactance are the same as those of resistance, and its SI unit is the ohm (\(\Omega\)).
In DC circuits \(f = 0\), and therefore \(X_L = 0\). This implies that a pure inductor offers zero resistance to DC, that is, it cannot reduce DC. Thus it passes DC and blocks AC of very high frequency. This is exactly the result you saw in the activity with the coil and the zigzag wire.
In an inductive circuit the self induced emf opposes the growth as well as the decay of current. It is not a one-way opposition: whichever way the current tries to change, the coil pushes back.
AC Voltage Applied to a Capacitor
Let us consider a capacitor of capacitance \(C\) connected to an AC source with an emf having the instantaneous value
\[e = e_0 \sin \omega t \qquad \text{--- (13.9)}\]
This is shown in Fig. 13.9.
Diagram
Fig. 13.9 – An AC source connected in a single loop to a capacitor drawn as two parallel plates.
The current flowing in the circuit transfers charge to the plates of the capacitor, which produces a potential difference between the plates. As the current reverses its direction in each half cycle, the capacitor is alternately charged and discharged. Note that no charge actually crosses the gap between the plates; the current in the wires exists because charge is being piled onto one plate and removed from the other, over and over.
Suppose \(q\) is the charge on the capacitor at any given instant \(t\). The potential difference across the plates of the capacitor is
\[V = \frac{q}{C} \qquad \text{or} \qquad q = CV \qquad \text{--- (13.10)}\]
where \(q\) is in coulomb (C), \(V\) in volt and \(C\) in farad (F).
The instantaneous value of the current \(i\) in the circuit is the rate at which charge arrives on the plates. Since the capacitor is connected straight across the source, \(V = e\) at every instant, so
\[i = \frac{dq}{dt} = \frac{d(Ce)}{dt} = C\frac{d(e_0 \sin \omega t)}{dt} = \omega C e_0 \cos \omega t\]
\[\therefore \; i = \frac{e_0 \cos \omega t}{\left(\dfrac{1}{\omega C}\right)}\]
Using the identity \(\cos \omega t = \sin\left(\omega t + \dfrac{\pi}{2}\right)\), this becomes
\[i = \frac{e_0}{\left(\dfrac{1}{\omega C}\right)} \sin\left(\omega t + \frac{\pi}{2}\right) \qquad \text{--- (13.11)}\]
The current will be maximum when \(\sin\left(\omega t + \dfrac{\pi}{2}\right) = 1\), so that \(i = i_0\), where the peak value of the current is
\[i_0 = \frac{e_0}{\left(\dfrac{1}{\omega C}\right)} \qquad \text{--- (13.12)}\]
\[\therefore \; i = i_0 \sin\left(\omega t + \frac{\pi}{2}\right) \qquad \text{--- (13.13)}\]
From Eq. (13.9) and Eq. (13.13) we find that in an AC circuit containing a capacitor only, the alternating current \(i\) leads the alternating emf \(e\) by a phase angle of \(\pi/2\) radian, as shown in Fig. 13.10. This is the exact opposite of what happened with the inductor, and the physical reason is easy to see: the current is largest when the capacitor is empty and easiest to charge, which is when the voltage across it is zero.
Diagram
Fig. 13.10 – Graph of e and i versus ωt for a pure capacitor: the current curve shifted a quarter cycle to the left of the emf curve, reaching its peak before the emf does.
Phasor Diagram for a Purely Capacitive Circuit
The phasor representing the peak emf makes an angle \(\omega t\) in the anticlockwise direction with respect to the horizontal axis. As the current leads the voltage by \(90^\circ\), the phasor representing the current \(i_0\) is turned \(90^\circ\) anticlockwise with respect to the phasor representing the emf \(e_0\), as shown in Fig. 13.11. The projections of these phasors on the vertical axis give the instantaneous values of \(e\) and \(i\).
Diagram
Fig. 13.11 – Phasor diagram for a purely capacitive circuit: the phasor \(e_0\) at angle ωt and the phasor \(i_0\) drawn 90° anticlockwise from it.
Capacitive Reactance
The instantaneous value of the alternating current through a capacitor is given by
\[i = \frac{e_0}{\left(\dfrac{1}{\omega C}\right)} \sin\left(\omega t + \frac{\pi}{2}\right) = i_0 \sin\left(\omega t + \frac{\pi}{2}\right)\]
Comparing Eq. (13.12) with Ohm’s law \(i_0 = \dfrac{e_0}{R}\), we find that the quantity \(\left(\dfrac{1}{\omega C}\right)\) represents the effective resistance offered by the capacitor. It is called the capacitive reactance and is denoted by \(X_C\).
\[\therefore \; X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}\]
where \(f\) is the frequency of the AC supply in hertz.
The function of capacitive reactance in a purely capacitive circuit is to limit the amplitude of the current, similar to the resistance in a purely resistive circuit.
\(X_C\) varies inversely as the frequency of the AC, and also inversely as the capacitance of the condenser. Both dependences make physical sense: a larger capacitance can absorb more charge for the same voltage, and a higher frequency gives the capacitor less time to fill up before the direction reverses, so in both cases more charge moves per second and the current is larger.
In a DC circuit, \(f = 0\), and therefore \(X_C = \infty\). Thus a capacitor blocks DC and acts as an open circuit, while it passes AC of high frequency.
The dimensions of capacitive reactance are the same as those of resistance, and its SI unit is the ohm (\(\Omega\)).
The differences between resistance and reactance are summarised in Table 13.1.
Table 13.1: Comparison between resistance and reactance.
| Resistance | Reactance |
|---|---|
| Equally effective for AC and DC. | Current is affected (reduced) but energy is not consumed (heat is not generated). The energy consumption by a coil is due to its resistive component. |
| Its value is independent of the frequency of the AC. | Inductive reactance \(\left(X_L = 2\pi f L\right)\) is directly proportional, and capacitive reactance \(\left(X_C = \dfrac{1}{2\pi f C}\right)\) is inversely proportional, to the frequency of the AC. |
| Current opposed by a resistor is in phase with the voltage. | Current opposed by a pure inductor lags in phase, while that opposed by a pure capacitor leads in phase, by \(\dfrac{\pi}{2}\) over the voltage. |
AC Circuit Containing Resistance, Inductance and Capacitance in Series (LCR Circuit)
Above we have studied the opposition offered by a resistor, a pure inductor and a capacitor to the flow of AC independently. Now let us consider the total opposition offered by a resistor, a pure inductor and a capacitor connected in series with an alternating source of emf, as shown in Fig. 13.12.
Diagram
Fig. 13.12 – Series LCR circuit: an AC source connected in one loop with a resistor R, an inductor L and a capacitor C, one after the other.
Let a pure resistor \(R\), a pure inductance \(L\) and an ideal capacitor of capacitance \(C\) be connected in series to a source of alternating emf. As \(R\), \(L\) and \(C\) are in series, the current at any instant through the three elements has the same amplitude and the same phase — this is the key simplification, and it is why the current, not the voltage, is taken as the reference. Let it be represented by
\[i = i_0 \sin \omega t\]
The voltage across each element bears a different phase relationship with this common current. The voltages \(e_R\), \(e_L\) and \(e_C\) are given by
\[e_R = iR, \qquad e_L = iX_L, \qquad e_C = iX_C\]
As the voltage across the capacitor lags behind the alternating current by \(90^\circ\), it is represented by \(\overrightarrow{OC}\), rotated clockwise through \(90^\circ\) from the direction of \(\vec{i_0}\). So \(\overrightarrow{OC}\) lies along OY\('\) in the phasor diagram shown in Fig. 13.13.
Diagram
Fig. 13.13 – Phasor diagram for an LCR circuit: \(e_R\) along OX, \(e_L\) along OY, \(e_C\) along OY′, the resultant reactive voltage \((e_L - e_C)\) marked OB′, and the diagonal OK of rectangle OAKB′ representing \(e_0\) at angle φ.
As \(e_R\) is in phase with the current \(i_0\), the vector \(e_R\) is drawn in the same direction as that of \(i\), along the positive direction of the X-axis, represented by \(\overrightarrow{OA}\).
The voltages across \(L\) and \(C\) have a phase difference of \(180^\circ\) between them — one points along OY and the other along OY\('\) — hence the net reactive voltage is \((e_L - e_C)\). Assuming \(e_L > e_C\), this is represented by OB\('\) in the figure.
The resultant of \(\overrightarrow{OA}\) and \(\overrightarrow{OB'}\) is the diagonal OK of the rectangle OAKB\('\). Since OA and OB\('\) are at right angles,
\[\therefore \; OK = \sqrt{OA^2 + OB'^2}\]
\[\therefore \; e_0 = \sqrt{e_R^2 + (e_L - e_C)^2} = \sqrt{(i_0 R)^2 + (i_0 X_L - i_0 X_C)^2}\]
\[\therefore \; e_0 = i_0 \sqrt{R^2 + (X_L - X_C)^2}\]
\[\therefore \; i_0 = \frac{e_0}{\sqrt{R^2 + (X_L - X_C)^2}} = \frac{e_0}{Z}\]
Comparing the above equation with the relation \(i = \dfrac{V}{R}\), we see that the quantity \(\sqrt{R^2 + (X_L - X_C)^2}\) represents the effective opposition offered by the inductor, capacitor and resistor connected in series to the flow of AC. This total effective resistance of an LCR circuit is called the impedance of the circuit and is represented by \(Z\).
The reciprocal of the impedance of an AC circuit is called admittance. Its SI unit is ohm\(^{-1}\) or siemens.
Phase Angle of an LCR Circuit
From the phasor diagram (Fig. 13.13) it can be seen that in an AC circuit containing \(L\), \(C\) and \(R\), the voltage leads the current by a phase angle \(\phi\), where
\[\tan \phi = \frac{AK}{OA} = \frac{OB'}{OA} = \frac{e_L - e_C}{e_R} = \frac{i_0 X_L - i_0 X_C}{i_0 R}\]
The common factor \(i_0\) cancels from numerator and denominator, which is why the phase angle depends only on the circuit elements and not on how large the current happens to be:
\[\therefore \; \tan \phi = \frac{X_L - X_C}{R} \qquad \therefore \; \phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)\]
Therefore the alternating current in an LCR circuit would be represented by
\[i = i_0 \sin (\omega t + \phi) \qquad \text{and} \qquad e = e_0 \sin (\omega t + \phi)\]
We can now discuss three cases based on the above discussion.
When \(X_L = X_C\), then \(\tan \phi = 0\). Hence the voltage and the current are in phase. Thus the AC circuit is non inductive.
When \(X_L > X_C\), \(\tan \phi\) is positive, and therefore \(\phi\) is positive. Hence the voltage leads the current by a phase angle \(\phi\). The AC circuit is an inductance dominated circuit.
When \(X_L < X_C\), \(\tan \phi\) is negative, and therefore \(\phi\) is negative. Hence the voltage lags the current by a phase angle \(\phi\). The AC circuit is a capacitance dominated circuit.
The first case is special, and it is the basis of everything you will study in Section 13.8: when the two reactances cancel exactly, the circuit behaves as though only the resistor were present.
Impedance Triangle
From the three phasors \(\vec{e_R} = i_0 R\), \(\vec{e_L} = i_0 X_L\) and \(\vec{e_C} = i_0 X_C\), we obtain the impedance triangle shown in Fig. 13.14. It is simply the voltage phasor diagram with every side divided by \(i_0\), which is allowed because the current is common to all three elements.
Diagram
Fig. 13.14 – Impedance triangle: a right-angled triangle with base R, perpendicular \((X_L - X_C)\), hypotenuse Z, and the angle φ between Z and R.
The diagonal OK represents the impedance \(Z\) of the AC circuit,
\[Z = \sqrt{R^2 + (X_L - X_C)^2}\]
the base OA represents the ohmic resistance \(R\), and the perpendicular AK represents the reactance \((X_L - X_C)\). The angle \(\angle AOK = \phi\) is the phase angle by which the voltage leads the current in the circuit, where
\[\tan \phi = \frac{X_L - X_C}{R}\]
This triangle is worth memorising as a picture rather than as a formula, because almost every LCR numerical asks either for its hypotenuse or for its angle.
13.6 Power in AC Circuit
We know that power is defined as the rate of doing work. For a DC circuit, power is measured simply as the product of voltage and current, because both stay fixed.
But in an AC circuit the values of current and voltage change at every instant. So the power in an AC circuit at a given instant is the product of the instantaneous voltage and the instantaneous current. This instantaneous power is itself a quantity that rises and falls many times a second, which is of little practical use — an electricity meter, a fuse or a heater responds to the average power over a full cycle. That average is what we now calculate, element by element.
Average Power Associated with Resistance
In a pure resistor the alternating current developed is in phase with the alternating voltage applied. That is, when
\[e = e_0 \sin \omega t \qquad \text{then} \qquad i = i_0 \sin \omega t\]
Now the instantaneous power is \(P = ei\), so
\[P = (e_0 \sin \omega t)(i_0 \sin \omega t) = e_0 i_0 \sin^2 \omega t \qquad \text{--- (13.14)}\]
Notice that \(\sin^2 \omega t\) can never be negative. The power is therefore positive at every instant, as shown in Fig. 13.15a: energy always flows from the source into the resistor and never returns.
Diagram
Fig. 13.15a – Two power-versus-time graphs side by side: for a resistor, a curve lying entirely above the time axis; for an inductor, a curve equally above and below the axis.
The instantaneous power varies with time, hence we consider the average power over a complete cycle by integrating Eq. (13.14) over one period \(T\) and dividing by \(T\):
\[P_{av} = \frac{\text{work done by the emf on the charges in one cycle}}{\text{time for one cycle}} = \frac{\int_0^T P \, dt}{T}\]
\[\therefore \; P_{av} = \frac{\int_0^T e_0 i_0 \sin^2 \omega t \; dt}{T} = \frac{e_0 i_0}{T} \int_0^T \sin^2 \omega t \; dt = \frac{e_0 i_0}{T}\left(\frac{T}{2}\right)\]
The integral equals \(T/2\) because the average value of \(\sin^2 \omega t\) over a complete cycle is \(\tfrac{1}{2}\), the same result that was used while deriving the rms value in Section 13.3. Hence
\[P_{av} = \frac{e_0 i_0}{2} = \left(\frac{e_0}{\sqrt{2}}\right)\left(\frac{i_0}{\sqrt{2}}\right) = e_{rms}\, i_{rms} \qquad \text{--- (13.15)}\]
This result is worth pausing on: the rms values are defined in precisely such a way that the AC power formula looks exactly like the DC one. \(P\) here is also called the apparent power, and it is measured in volt-ampere (VA).
Average Power Associated with an Inductor
In a purely inductive circuit the current lags behind the voltage by a phase angle of \(\pi/2\). That is, when
\[e = e_0 \sin \omega t \qquad \text{then} \qquad i = i_0 \sin\left(\omega t - \frac{\pi}{2}\right)\]
Now the instantaneous power is \(P = ei\):
\[P = (e_0 \sin \omega t)\, i_0 \sin\left(\omega t - \frac{\pi}{2}\right) = -\, e_0 i_0 \sin \omega t \cos \omega t\]
where \(\sin\left(\omega t - \dfrac{\pi}{2}\right) = -\cos \omega t\) has been used. Averaging over a complete cycle,
\[P_{av} = \frac{\int_0^T P \, dt}{T} = -\frac{e_0 i_0}{T}\int_0^T \sin \omega t \cos \omega t \; dt = -\frac{e_0 i_0}{2T}\int_0^T \sin 2\omega t \; dt = 0\]
The last integral vanishes because \(\sin 2\omega t\) completes two full oscillations in one period \(T\), so its positive and negative parts cancel exactly.
Therefore the average power over a complete cycle of AC through an ideal inductor is zero.
Average Power Associated with a Capacitor
In a purely capacitive circuit the current leads the emf by a phase angle of \(\pi/2\). That is, when
\[e = e_0 \sin \omega t \qquad \text{then} \qquad i = i_0 \sin\left(\omega t + \frac{\pi}{2}\right) = i_0 \cos \omega t\]
Now the instantaneous power is
\[P = ei = (e_0 \sin \omega t)(i_0 \cos \omega t) = e_0 i_0 \sin \omega t \cos \omega t\]
\[\therefore \; P_{av} = \frac{\int_0^T P \, dt}{T} = \frac{e_0 i_0}{T}\int_0^T \sin \omega t \cos \omega t \; dt = 0\]
as shown above. Hence the average power supplied to an ideal capacitor by the source over a complete cycle of AC is also zero. Here the energy is stored in the electric field between the plates during charging and handed back during discharging.
Average Power in an LCR Circuit
Let \(e = e_0 \sin \omega t\) be the alternating emf applied across the series combination of a pure inductor, a capacitor and a resistor, as shown in Fig. 13.15.
Diagram
Fig. 13.15 – LCR series circuit connected to an AC source, with the phase angle φ between applied emf and current indicated.
There is a phase difference \(\phi\) between the applied emf and the current, so that
\[i = i_0 \sin (\omega t \pm \phi)\]
The instantaneous power is given by
\[P = ei = (e_0 \sin \omega t)\, i_0 \sin (\omega t \pm \phi)\]
Expanding the second sine using \(\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B\),
\[P = e_0 i_0 \left[\sin \omega t \cos \phi \pm \cos \omega t \sin \phi\right] \sin \omega t\]
\[P = e_0 i_0 \left[\sin^2 \omega t \cos \phi \pm \cos \omega t \sin \phi \sin \omega t\right]\]
Therefore the average power is
\[P_{av} = \frac{\text{work done in one cycle}}{\text{time for one cycle}} = \frac{\int_0^T P \, dt}{T}\]
\[P_{av} = \frac{e_0 i_0}{T}\left[\cos \phi \int_0^T \sin^2 \omega t \; dt \; \pm \; \sin \phi \int_0^T \sin \omega t \cos \omega t \; dt \right] \qquad \text{--- (13.16)}\]
As seen above,
\[\int_0^T \sin^2 \omega t \; dt = \frac{T}{2} \qquad \text{and} \qquad \int_0^T \sin \omega t \cos \omega t \; dt = 0 \qquad \text{--- (13.17)}\]
Substituting Eq. (13.17) in Eq. (13.16), the entire second term disappears and only the first survives:
\[P_{av} = \frac{e_0 i_0}{T} \cdot \frac{T}{2} \cos \phi = \left(\frac{e_0}{\sqrt{2}}\right)\left(\frac{i_0}{\sqrt{2}}\right)\cos \phi\]
\[\therefore \; P_{av} = e_{rms}\, i_{rms} \cos \phi \qquad \text{--- (13.18)}\]
This power \(P_{av}\) is also called the true power, and it is measured in watt (W). The average power dissipated in an AC circuit of an inductor, capacitor and resistor connected in series depends not only on the rms values of current and emf but also on the phase difference \(\phi\) between them.
Power Factor
The factor \(\cos \phi\) is called the power factor. It measures what fraction of the apparent power is actually converted into useful work or heat.
\[\text{Power factor } (\cos \phi) = \frac{\text{true power } (P_{av})}{\text{apparent power } (P)} = \frac{R}{\sqrt{R^2 + (X_L - X_C)^2}}\]
The last step comes straight from the impedance triangle of Fig. 13.14, in which \(R\) is the base and \(Z\) the hypotenuse. Hence
\[\therefore \; \text{Power factor } \cos \phi = \frac{\text{resistance}}{\text{impedance}} = \frac{R}{Z}\]
In a non inductive circuit, \(X_L = X_C\), so
\[\text{Power factor } (\cos \phi) = \frac{R}{\sqrt{R^2}} = \frac{R}{R} = 1, \qquad \phi = 0\]
In a purely inductive or purely capacitive circuit, \(\phi = 90^\circ\), and therefore the power factor is zero. The average power consumed in a pure inductor or an ideal capacitor is
\[P_{av} = e_{rms}\, i_{rms} \cos 90^\circ = \text{zero}\]
Hence the current through a pure inductor or an ideal capacitor, which consumes no power for its maintenance in the circuit, is called idle current or wattless current. Power dissipated in a circuit is due to resistance only.
In numerical problems, the most common mistake is to write \(P = e_{rms} i_{rms}\) for a circuit that contains an inductor or a capacitor. Always check whether the circuit has a phase difference; if it does, the factor \(\cos \phi\) must be included, and it is often easier to find it as \(R/Z\) than by first calculating the angle.
13.7 LC Oscillations
We have seen in Chapters 8, 10 and 12 that capacitors and inductors store energy in their electric and magnetic fields respectively. A capacitor stores energy in the electric field between its plates, and an inductor stores it in the magnetic field around its turns. What happens if we connect one to the other and leave them alone?
When a capacitor is supplied with an AC current it gets charged. When such a fully charged capacitor is connected to an inductor, the charge is transferred to the inductor and current starts flowing through the inductor.
Because of the increasing current there will be a change in the magnetic flux of the inductor in the circuit. Hence an induced emf is produced in the circuit. This self-induced emf will try to oppose the growth of the current.
Due to this, the charge on the capacitor — and therefore the energy stored in it — decreases, and an equivalent amount of energy is stored in the inductor in the form of a magnetic field. When the discharging of the capacitor is complete, all the energy that was stored in the capacitor will be stored in the inductor. The capacitor will have become fully discharged, whereas the inductor will be storing all the energy.
As a result, the inductor will now start charging the capacitor. The current and the magnetic flux linked with the inductor start decreasing, therefore an induced emf is produced which recharges the capacitor in the opposite direction. This process of charging and discharging of the capacitor is repeated, and the energy taken from the source keeps on oscillating between the capacitor \(C\) and the inductor \(L\).
The complete sequence is explained stage by stage in Fig. 13.16.
One Complete Cycle, Stage by Stage
Let a capacitor with initial charge \(q_0\) at \(t = 0\) be connected to an ideal inductor of zero resistance, as shown in Fig. 13.16 (a). The electrical energy stored in the dielectric medium between the plates of the capacitor is
\[U_E = \frac{1}{2}\frac{q_0^{2}}{C}\]
Since there is no current in the circuit at this instant, the energy stored in the magnetic field of the inductor is zero.
Diagram
Fig. 13.16 (a) – A capacitor carrying charge \(q_0\) connected through an open key to an ideal inductor; all energy marked as electric.
As the circuit is closed, the capacitor begins to discharge itself through the inductor, giving rise to a current \(I\) in the circuit, as shown in Fig. 13.16 (b). As the current \(I\) increases, it builds up a magnetic field around the inductor. A part of the electrical energy of the capacitor gets stored in the inductor in the form of magnetic energy
\[U_B = \frac{1}{2} L I^{2}\]
where \(L\) is the inductance in henry (H) and \(I\) is the instantaneous current in ampere (A).
Diagram
Fig. 13.16 (b) – The same circuit with the key closed, a partly discharged capacitor, current I flowing, and magnetic field lines forming around the coil.
At a later instant the capacitor gets fully discharged and the potential difference across its plates becomes zero, as in Fig. 13.16 (c). At this moment the current reaches its maximum value \(I_0\), and the energy in the magnetic field is \(\dfrac{1}{2} L I_0^{2}\). Thus the entire electrostatic energy of the capacitor has been converted into the magnetic field energy of the inductor.
Diagram
Fig. 13.16 (c) – Capacitor plates uncharged, current at its maximum value \(I_0\), all energy marked as magnetic.
At first sight one might expect the current to stop here, since there is no longer any voltage to drive it. It does not, and the reason is the inductor. After the discharge of the capacitor is complete, the magnetic flux linked with the inductor decreases, inducing a current in the same direction as the earlier current, in accordance with Lenz’s law. The current thus persists, but with decreasing magnitude, and charges the capacitor in the opposite direction, as shown in Fig. 13.16 (d). The magnetic energy of the inductor begins to change into the electrostatic energy of the capacitor.
Diagram
Fig. 13.16 (d) – Current continuing in the same direction but decreasing, charging the capacitor with reversed polarity.
The process continues till the capacitor is fully charged with a polarity which is opposite to that in its initial state, as shown in Fig. 13.16 (e). Thus the entire energy is again stored as \(\dfrac{1}{2}\dfrac{q_0^{2}}{C}\) in the electric field of the capacitor.
Diagram
Fig. 13.16 (e) – Capacitor fully charged with polarity opposite to Fig. 13.16 (a), zero current in the circuit.
The capacitor begins to discharge again, sending current in the opposite direction. The energy is once again transferred to the magnetic field of the inductor. Thus the process repeats itself in the opposite direction, and the circuit eventually returns to its initial state. Only then is one full cycle complete.
Energy Surging Back and Forth
Thus the energy of the system continuously surges back and forth between the electric field of the capacitor and the magnetic field of the inductor. This produces electrical oscillations of a definite frequency. These are called LC oscillations. If there is no loss of energy, the amplitude of the oscillations remains constant and the oscillations are undamped.
Because the exchange of energy is complete in an ideal circuit, the two peak energies must be equal:
\[\frac{1}{2}\frac{q_0^{2}}{C} = \frac{1}{2} L I_0^{2}\]
This single statement of energy conservation is enough to find the maximum current in an LC circuit whenever the initial charge is known, and it is a favourite starting point for numerical questions on this section. The rate at which the energy sloshes to and fro is fixed by \(L\) and \(C\) alone; you will see in Section 13.8 that this natural frequency is \(f_r = \dfrac{1}{2\pi\sqrt{LC}}\).
Why LC Oscillations Die Out in Practice
However, LC oscillations are usually damped, due to the following reasons.
Every inductor has some resistance. This causes energy loss as heat. The amplitude of the oscillations goes on decreasing and they finally die out.
Even if the resistance were zero, the total energy of the system would not remain constant. It is radiated away in the form of electromagnetic waves. The working of radio and TV transmitters is based on such radiations.
13.8 Electric Resonance
Have you ever wondered how a radio picks out certain frequencies so that you can play your favourite channel, or why a glass can break during an orchestra concert? Why do you think you encounter such situations? The answer to both lies in the phenomenon of resonance.
The phenomenon of resonance can be observed in systems that have a tendency to oscillate at a particular frequency, which is called the natural frequency of oscillation of the system. When such a system is driven by an energy source whose frequency is equal to the natural frequency of the system, the amplitude of the oscillations becomes large, and resonance is said to occur.
An LCR circuit, as we saw in the previous section, has its own natural frequency of oscillation fixed by \(L\) and \(C\). So we should expect something dramatic to happen when the frequency of the AC source is made equal to it.
Series Resonance Circuit
A circuit in which inductance \(L\), capacitance \(C\) and resistance \(R\) are connected in series, as in Fig. 13.17, and which admits maximum current corresponding to a given frequency of AC, is called a series resonance circuit.
Diagram
Fig. 13.17 – Series resonance circuit: L, C and R in series with a variable-frequency AC source and an ammeter.
The impedance \(Z\) of an LCR circuit is given by
\[Z = \sqrt{R^{2} + \left(\omega L - \frac{1}{\omega C}\right)^{2}}\]
At very low frequencies the inductive reactance \(X_L = \omega L\) is negligible, but the capacitive reactance \(X_C = \dfrac{1}{\omega C}\) is very high. As we increase the applied frequency, \(X_L\) increases and \(X_C\) decreases, as shown in Fig. 13.17a. Since one is rising and the other is falling, there must be one frequency at which they are exactly equal.
Diagram
Fig. 13.17a – Graph of reactance versus frequency: a straight line rising for \(X_L\) and a falling curve for \(X_C\), crossing at the resonant frequency \(f_r\).
At some angular frequency \(\omega_r\), therefore, \(X_L = X_C\):
\[\omega_r L = \frac{1}{\omega_r C}\]
\[\therefore \; (\omega_r)^{2} = \frac{1}{LC} \qquad \text{or} \qquad 4\pi^{2} f_r^{2} = \frac{1}{LC}\]
\[\therefore \; 2\pi f_r = \frac{1}{\sqrt{LC}}\]
\[\therefore \; f_r = \frac{1}{2\pi \sqrt{LC}}\]
where \(f_r\) is called the resonant frequency, measured in hertz, with \(L\) in henry and \(C\) in farad. Notice that it does not depend on \(R\) at all.
At this particular frequency \(f_r\), since \(X_L = X_C\), we get
\[Z = \sqrt{R^{2} + 0} = R\]
This is the least value of \(Z\). Thus, when the impedance of an LCR circuit is minimum, the circuit is said to be purely resistive, the current and voltage are in phase, and hence the current
\[i_0 = \frac{e_0}{Z} = \frac{e_0}{R}\]
is maximum. This condition of the LCR circuit is called the resonance condition, and this frequency is called the series resonant frequency.
At \(\omega = \omega_r\) the value of the peak current \(i_0\) is maximum. This maximum value of the peak current is inversely proportional to \(R\), since \(i_0 = \dfrac{e_0}{R}\). For lower values of \(R\), \(i_0\) is large, and vice versa. The variation of rms current with the frequency of the AC is shown in the graph of Fig. 13.18. The curve is called the series resonance curve, and at resonance the rms current becomes maximum.
Diagram
Fig. 13.18 – Series resonance curve: rms current plotted against frequency, showing a sharp peak at \(f_r\), with a second flatter curve drawn for a larger value of R.
This circuit at the resonant condition is very useful for radio and TV receivers for tuning in the signal from a desired transmitting station or channel.
Characteristics of a Series Resonance Circuit
- Resonance occurs when \(X_L = X_C\).
- The resonant frequency is \(f_r = \dfrac{1}{2\pi\sqrt{LC}}\).
- Impedance is minimum and the circuit is purely resistive.
- Current has a maximum value.
- When a number of frequencies are fed to it, it accepts only one frequency \(f_r\) and rejects the other frequencies. The current is maximum for this frequency. Hence it is called an acceptor circuit.
Parallel Resonance Circuit
A parallel resonance circuit consists of a coil of inductance \(L\) and a condenser of capacity \(C\) joined in parallel to a source of alternating emf, as shown in Fig. 13.19.
Diagram
Fig. 13.19 – Parallel resonance circuit: an inductor L and a capacitor C connected as two parallel branches across an AC source.
Let the alternating emf supplied by the source be
\[e = e_0 \sin \omega t\]
Since the two elements are in parallel, the same voltage appears across both, but each carries its own current with its own phase. In the case of an inductor the current lags behind the applied emf by a phase angle of \(\pi/2\), so the instantaneous current through \(L\) is given by
\[i_L = \frac{e_0}{X_L}\sin\left(\omega t - \pi/2\right)\]
Similarly, in a capacitor the current leads the emf by a phase angle of \(\pi/2\), so we can write
\[i_C = \frac{e_0}{X_C}\sin\left(\omega t + \pi/2\right)\]
Therefore the total current \(i\) in the circuit at this instant is
\[i = i_C + i_L = \frac{e_0}{X_L}\sin\left(\omega t - \pi/2\right) + \frac{e_0}{X_C}\sin\left(\omega t + \pi/2\right)\]
\[= \frac{e_0}{X_L}(-\cos \omega t) + \frac{e_0}{X_C}\cos \omega t\]
\[= e_0 \cos \omega t \left(\frac{1}{X_C} - \frac{1}{X_L}\right)\]
\[\therefore \; i = e_0 \cos \omega t \left(\omega C - \frac{1}{\omega L}\right)\]
The two branch currents are half a cycle apart, so they flow in opposite directions in the main line and subtract from each other. We find that \(i\) is minimum when
\[\omega C - \frac{1}{\omega L} = 0\]
\[\text{i.e.} \quad \omega C = \frac{1}{\omega L} \quad \text{i.e.} \quad \omega^{2} = \frac{1}{LC}\]
\[\therefore \; \omega = \frac{1}{\sqrt{LC}} \qquad \text{or} \qquad 2\pi f_r = \frac{1}{\sqrt{LC}}\]
\[\therefore \; f_r = \frac{1}{2\pi \sqrt{LC}}\]
where \(f_r\) is again called the resonant frequency. It has the same form as for the series circuit, but its consequences are exactly opposite.
Therefore at the parallel resonance frequency \(f_r\), the current \(i\) is minimum, that is, the circuit allows minimum current to flow through it, as shown in the graph of Fig. 13.20. The impedance is maximum at this frequency. Such a circuit is called a parallel resonance circuit. A parallel resonant circuit is very useful in wireless transmission or radio communication and in filter circuits.
Diagram
Fig. 13.20 – Parallel resonant curve: current plotted against frequency, dipping to a minimum value \(i_{min}\) at \(f_r\).
Characteristics of a Parallel Resonance Circuit
- Resonance occurs when \(X_L = X_C\).
- The resonant frequency is \(f_r = \dfrac{1}{2\pi\sqrt{LC}}\).
- Impedance is maximum.
- Current is minimum.
- When alternating currents of different frequencies are sent through a parallel resonant circuit, it offers a very high impedance to the current of the resonant frequency \(f_r\) and rejects it, but allows the currents of other frequencies to pass through it. Hence it is called a rejector circuit.
13.10 Choke Coil
Suppose a device needs less current than the mains supply would push through it. The obvious solution is to put a resistance in series with it. But if we use a resistance to reduce the current passing through an AC circuit, there will be a loss of electric energy in the form of heat, given by \(I^{2}RT\), due to Joule heating. That energy is simply wasted, and the resistor also becomes hot. A choke coil helps to minimise this effect.
Construction
A choke coil is made up of thick insulated copper wires wound closely in a large number of turns over a soft iron laminated core, as shown in Fig. 13.22.
Diagram
Fig. 13.22 – A choke coil: many closely wound turns of thick insulated copper wire on a laminated soft iron core, shown connected in series with a lamp and an AC source.
Every part of this description has a purpose. A large number of turns wound on an iron core makes the self inductance \(L\) very large, and thick copper wire keeps the ohmic resistance \(R\) of the winding very small.
The choke coil offers a large reactance \(X_L = \omega L\) to the flow of AC, and hence the current is reduced. The laminated core reduces eddy current losses, since the thin insulated sheets do not allow large circulating currents to be set up inside the iron.
Why Almost No Energy Is Lost
The average power dissipated in the choke is
\[P = I_{rms} E_{rms} \cos \phi\]
where the power factor is
\[\cos \phi = \frac{R}{\sqrt{R^{2} + \omega^{2}L^{2}}}\]
For a choke coil, \(L\) is very large while \(R\) is very small. The denominator is therefore far bigger than the numerator, so \(\cos \phi\) is nearly zero and the power loss is very small. The current is cut down by the reactance, which consumes no power, instead of by resistance, which does.
The only loss of energy is due to hysteresis loss in the iron core, which can be reduced by using a soft iron core, since soft iron is easy to magnetise and demagnetise repeatedly.
13.11 Summary
Alternating Current and Its Values
In a DC supply the polarity of the voltage is fixed; in an AC supply the polarity keeps changing periodically, so the current reverses direction again and again.
An AC generator produces a time dependent emf \(e = e_0 \sin \omega t\), where \(e_0\) is the peak emf in volt and \(\omega\) is the angular frequency in rad s\(^{-1}\). The resulting current is \(i = i_0 \sin(\omega t + \alpha)\), where \(\alpha\) is the phase difference between the current and the emf.
The direction of the emf reverses after every half revolution of the generator coil. Such an emf is an alternating emf and the current it drives is an alternating current.
Peak value is the maximum value of the alternating current (or emf) in either direction, written as \(i_0\) or \(e_0\).
The average value over a full cycle is zero, so the average value of AC is defined over one half cycle: \[i_{av} = 0.637\, i_0 \qquad e_{av} = 0.637\, e_0\]
The rms value is the steady current that would produce the same heat in the same resistance in the same time: \[i_{rms} = \frac{i_0}{\sqrt{2}} = 0.707\, i_0 \qquad e_{rms} = \frac{e_0}{\sqrt{2}} = 0.707\, e_0\]
Heat produced over a complete cycle: \(H = i_{rms}^{2} R t\). AC ammeters and voltmeters read rms values.
Phasors
A rotating vector representing a sinusoidally varying quantity is called a phasor; the diagram containing such vectors is a phasor diagram.
The length of the phasor gives the peak value, its angle with the horizontal axis is \(\omega t\) or \(\omega t + \alpha\), it rotates anticlockwise, and its projection on a fixed axis gives the instantaneous value. Phasors allow alternating quantities to be added by the laws of vector addition.
Single Elements in an AC Circuit
Resistor: \(e = e_0 \sin \omega t\) gives \(i = i_0 \sin \omega t\) with \(i_0 = \dfrac{e_0}{R}\). Current and voltage are in phase. \(R\) behaves the same for AC and DC.
Inductor: \(i = i_0 \sin\left(\omega t - \dfrac{\pi}{2}\right)\) with \(i_0 = \dfrac{e_0}{\omega L}\). The current lags the emf by \(\dfrac{\pi}{2}\) radian (\(90^\circ\)). \[X_L = \omega L = 2\pi f L\] Inductive reactance is measured in ohm, is proportional to \(L\) and \(f\), and is zero for DC (\(f = 0\)), so an inductor passes DC and blocks high-frequency AC.
Capacitor: \(i = i_0 \sin\left(\omega t + \dfrac{\pi}{2}\right)\) with \(i_0 = \dfrac{e_0}{\left(1/\omega C\right)}\). The current leads the emf by \(\dfrac{\pi}{2}\) radian. \[X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}\] Capacitive reactance is measured in ohm, varies inversely with \(f\) and \(C\), and is infinite for DC, so a capacitor blocks DC and passes high-frequency AC.
Reactance reduces the current but consumes no energy; resistance both reduces the current and dissipates heat.
Series LCR Circuit
With \(R\), \(L\) and \(C\) in series, the current is common to all three, and \[e_R = iR, \qquad e_L = iX_L, \qquad e_C = iX_C\]
Impedance: \[Z = \sqrt{R^{2} + (X_L - X_C)^{2}} \qquad i_0 = \frac{e_0}{Z}\] Impedance is expressed in ohm. Its reciprocal is called admittance, with SI unit ohm\(^{-1}\) or siemens.
Phase angle: \[\tan \phi = \frac{X_L - X_C}{R} \qquad \phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)\]
If \(X_L = X_C\), the circuit is non inductive and \(e\) and \(i\) are in phase. If \(X_L > X_C\), the voltage leads the current and the circuit is inductance dominated. If \(X_L < X_C\), the voltage lags the current and the circuit is capacitance dominated.
In the impedance triangle, the base is \(R\), the perpendicular is \((X_L - X_C)\), the hypotenuse is \(Z\), and the angle between \(Z\) and \(R\) is \(\phi\).
Power in an AC Circuit
Instantaneous power is the product of instantaneous voltage and instantaneous current.
For a pure resistor: \(P_{av} = \dfrac{e_0 i_0}{2} = e_{rms}\, i_{rms}\), called the apparent power.
For an ideal inductor and for an ideal capacitor, the average power over a complete cycle is zero; energy is stored and returned, not dissipated.
For a series LCR circuit, the true power is \[P_{av} = e_{rms}\, i_{rms} \cos \phi\]
The power factor is \[\cos \phi = \frac{\text{true power}}{\text{apparent power}} = \frac{R}{Z}\] It is 1 for a non inductive circuit and 0 for a purely inductive or purely capacitive circuit.
A current that consumes no average power, being \(90^\circ\) out of phase with the emf, is called a wattless or idle current. Power is dissipated only in resistance.
LC Oscillations
A charged capacitor discharging through a non-resistive inductor produces electrical oscillations of constant amplitude and frequency, called LC oscillations.
The energy surges back and forth between the electric field of the capacitor, \(U_E = \dfrac{1}{2}\dfrac{q_0^{2}}{C}\), and the magnetic field of the inductor, \(U_B = \dfrac{1}{2}LI^{2}\), reaching \(\dfrac{1}{2}LI_0^{2}\) when the capacitor is fully discharged.
In practice the oscillations are damped, because every inductor has some resistance which dissipates heat, and because energy is radiated away as electromagnetic waves. Radio and TV transmitters make use of such radiation.
Electric Resonance
Resonance occurs when a system is driven at its own natural frequency, making the amplitude of oscillation maximum.
Resonance in an LCR circuit occurs when \(X_L = X_C\), at the resonant frequency \[f_r = \frac{1}{2\pi\sqrt{LC}} \qquad \omega_r = \frac{1}{\sqrt{LC}}\] Both \(L\) and \(C\) must be present; there is no resonance in an LR or a CR circuit.
Series resonance: impedance is minimum and equal to \(R\), the circuit is purely resistive, current is maximum at \(i_0 = \dfrac{e_0}{R}\), and the circuit accepts one frequency and rejects others — an acceptor circuit, used for tuning radio and TV receivers.
Parallel resonance: impedance is maximum and current is minimum; the circuit rejects the resonant frequency and passes others — a rejector circuit, used in wireless communication and filter circuits.
Choke Coil
A choke coil is an inductor used to reduce AC in a circuit without much loss of energy, in place of a resistor which would waste energy as \(I^{2}RT\).
It has a large number of turns of thick insulated copper wire on a laminated soft iron core: many turns make \(L\) large, thick wire keeps \(R\) small, lamination reduces eddy current loss and soft iron reduces hysteresis loss.
Power dissipated is \(P = I_{rms} E_{rms}\cos\phi\) with \(\cos\phi = \dfrac{R}{\sqrt{R^{2} + \omega^{2}L^{2}}}\). Since \(L\) is large and \(R\) is small, \(\cos\phi\) is nearly zero and the power loss is very small.
13.12 Objective Questions
- Choose the correct option.
- If the rms current in a 50 Hz AC circuit is 5 A, the value of the current \(\dfrac{1}{300}\) seconds after its value becomes zero is
- \(5\sqrt{2}\) A
- \(5\sqrt{\dfrac{3}{2}}\) A
- \(\dfrac{5}{6}\) A
- \(\dfrac{5}{\sqrt{2}}\) A
- A resistor of 500 \(\Omega\) and an inductance of 0.5 H are in series with an AC source which is given by \(V = 100\sqrt{2}\,\sin(1000\,t)\). The power factor of the combination is
- \(\dfrac{1}{\sqrt{2}}\)
- \(\dfrac{1}{\sqrt{3}}\)
- 0.5
- 0.6
- In a circuit L, C and R are connected in series with an alternating voltage of frequency \(f\). The current leads the voltage by \(45^\circ\). The value of C is
- \(\dfrac{1}{\pi f\left(2\pi f L - R\right)}\)
- \(\dfrac{1}{2\pi f\left(2\pi f L - R\right)}\)
- \(\dfrac{1}{\pi f\left(2\pi f L + R\right)}\)
- \(\dfrac{1}{2\pi f\left(2\pi f L + R\right)}\)
- In an AC circuit, e and i are given by \(e = 150 \sin(150\,t)\) V and \(i = 150 \sin\left(150\,t + \dfrac{\pi}{3}\right)\) A. The power dissipated in the circuit is
- 106 W
- 150 W
- 5625 W
- Zero
- In a series LCR circuit the phase difference between the voltage and the current is \(45^\circ\). Then the power factor will be
- 0.607
- 0.707
- 0.808
- 1
13.13 Check Your Concepts
Why is the mean value of an alternating current defined over half a cycle and not over a full cycle? What would be the value if a full cycle were used, and why is that value of no practical use?
A moving coil ammeter shows no deflection when connected in an AC circuit, although the circuit clearly carries a current. Explain this observation, and state which property of the current is used by an AC ammeter instead.
Two identical heating elements are connected, one to a 220 V DC supply and the other to a 220 V rms AC supply of 50 Hz. Will they produce the same amount of heat in the same time? Justify your answer.
Explain physically, without using the formula, why the current through a pure inductor lags the applied emf by a quarter of a cycle while the current through a pure capacitor leads it by a quarter of a cycle.
A lamp is connected in series with a capacitor to an AC source. What happens to the brightness of the lamp if (a) a dielectric slab is inserted between the plates of the capacitor, and (b) the frequency of the source is reduced? Give reasons.
In a series LCR circuit at resonance, a voltmeter placed across the inductor may read a voltage larger than the emf of the source. Does this violate conservation of energy? Explain.
Why is it not possible to obtain resonance in a circuit containing only L and R, or only C and R? What is the essential role played by having both L and C present?
A series resonant circuit is called an acceptor circuit while a parallel resonant circuit is called a rejector circuit. Compare the two in terms of impedance and current at the resonant frequency, and explain how each name is justified.
What is meant by wattless current? Identify the circuit elements that carry such a current and explain where the energy drawn from the source goes during each cycle.
An engineer needs to reduce the current in an AC circuit. Explain why a choke coil is preferred over a rheostat, and state what would happen if the same choke coil were used in a DC circuit instead.
Why is the core of a choke coil made of soft iron, and why is it laminated rather than solid? Name the energy loss each of these two design choices reduces.
Two series LCR circuits have the same values of L and C but different resistances. Which of them has the sharper resonance curve, which has the larger Q factor, and which would be more suitable for use in a radio tuner? Explain your reasoning.
Explain why LC oscillations in a real circuit are damped, even though the idealised treatment predicts oscillations of constant amplitude. State both reasons and identify which one is put to practical use.
Assertion: In a purely capacitive AC circuit, the average power consumed over a complete cycle is zero. Reason: The current in such a circuit is zero at every instant. State whether the assertion and the reason are individually correct, and whether the reason correctly explains the assertion.
A factory operating many motors has a power factor of 0.6. Explain why the electricity supply company objects to this, and describe what can be connected across the supply to improve the situation and why it works.
Below its resonant frequency a series LCR circuit behaves as a capacitance dominated circuit, and above it as an inductance dominated circuit. Explain why, and state what happens to the phase angle \(\phi\) as the frequency is increased from well below resonance to well above it.
Answer in brief.
- An electric lamp is connected in series with a capacitor and an AC source is glowing with a certain brightness. How does the brightness of the lamp change on increasing the capacitance?
- The total impedance of a circuit decreases when a capacitor is added in series with L and R. Explain why.
- For very high frequency AC supply, a capacitor behaves like a pure conductor. Why?
- What is wattless current?
- What is the natural frequency of an LC circuit? What is the reactance of this circuit at this frequency?
When an AC source is connected to an ideal inductor, show that the average power supplied by the source over a complete cycle is zero.
Prove that an ideal capacitor in an AC circuit does not dissipate power.
An emf \(e = e_0 \sin \omega t\) applied to a series L–C–R circuit derives a current \(I = I_0 \sin \omega t\) in the circuit. Deduce the expression for the average power dissipated in the circuit.
For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
A device Y is connected across an AC source of emf \(e = e_0 \sin \omega t\). The current through Y is given as \(i = i_0 \sin\left(\omega t + \dfrac{\pi}{2}\right)\).
- Identify the device Y and write the expression for its reactance.
- Draw graphs showing variation of emf and current with time over one cycle of AC for Y.
- How does the reactance of the device Y vary with the frequency of the AC? Show graphically.
- Draw the phasor diagram for the device Y.
Derive an expression for the impedance of an LCR circuit connected to an AC power supply.
Compare resistance and reactance.
Show that in an AC circuit containing a pure inductor, the voltage is ahead of the current by \(\dfrac{\pi}{2}\) in phase.
An AC source generating a voltage \(e = e_0 \sin \omega t\) is connected to a capacitor of capacitance C. Find the expression for the current \(i\) flowing through it. Plot a graph of \(e\) and \(i\) versus \(\omega t\).
13.14 Practice with Numericals
An alternating voltage is given by \(e = 311 \sin(314\,t)\) volt. Find (i) the peak value of the voltage, (ii) its rms value, (iii) the frequency, (iv) the time period, and (v) the instantaneous voltage at \(t = 5\) ms.
An AC ammeter in a circuit reads 3.0 A when connected in series with a resistor of 60 \(\Omega\). Calculate the peak value of the current, the rms voltage across the resistor, and the average power dissipated in it.
A coil of inductance 0.5 H and negligible resistance is connected across a 200 V, 50 Hz supply. Calculate its inductive reactance and the rms current. If the frequency of the supply is raised to 200 Hz at the same voltage, find the new reactance and current.
A 20 \(\mu\)F capacitor is connected across a 220 V, 50 Hz supply. Calculate the capacitive reactance, the rms current and the peak current in the circuit.
A resistor of 30 \(\Omega\), an inductor of 0.2 H and a capacitor of 50 \(\mu\)F are connected in series across a 230 V, 50 Hz supply. Calculate (i) \(X_L\), (ii) \(X_C\), (iii) the impedance, (iv) the rms current, (v) the phase angle, and (vi) the power factor of the circuit.
For the circuit of the previous problem, calculate the true power and the apparent power, and state how much power would have been calculated if the power factor had been wrongly ignored.
A series LCR circuit has \(L = 0.4\) H, \(C = 25\ \mu\)F and \(R = 10\ \Omega\), and is connected to a 100 V variable-frequency source. Calculate the resonant frequency, the impedance at resonance and the current at resonance.
A series resonant circuit has a resonant frequency of 800 kHz and a bandwidth of 20 kHz. Calculate its Q factor. If the resistance of the circuit is doubled without changing L and C, state qualitatively what happens to the resonant frequency and to the bandwidth.
A choke coil of inductance 0.8 H and resistance 6 \(\Omega\) is connected to a 230 V, 50 Hz supply. Calculate its impedance, the rms current, the power factor and the average power dissipated in the coil.
A capacitor of 25 \(\mu\)F is charged to a potential difference of 20 V and then connected across a pure inductor of inductance 10 mH. Calculate the maximum current in the circuit and the frequency of the LC oscillations produced.
Find the time taken by a 50 Hz alternating current to change from zero to its peak value. Also find the time it takes to change from its peak value to half of the peak value.
What capacitance must be connected in series with an inductance of 50 mH so that the combination resonates at 1.0 kHz?
An electric lamp is rated 60 W, 220 V and is run on a 220 V, 50 Hz supply. Assuming it behaves as a pure resistance, calculate its resistance, the rms current through it and the peak current through it.
In an AC circuit the applied emf and the resulting current are \(e = 100 \sin(314\,t)\) V and \(i = 5 \sin\left(314\,t - \dfrac{\pi}{6}\right)\) A. Calculate the rms values of emf and current, the power factor and the average power dissipated in the circuit.
An alternating current has a peak value of 8 A. Calculate its rms value and its average value over a half cycle. Also find the heat produced in a 20 \(\Omega\) resistor by this current in 2 minutes.
A series circuit has a resistance of 40 \(\Omega\) and an inductive reactance of 30 \(\Omega\) at 50 Hz. Calculate the impedance and the power factor. What value of capacitive reactance must be introduced in series so that the power factor of the circuit becomes unity, and what capacitance does this correspond to?
In a series LR circuit \(X_L = R\) and the power factor of the circuit is \(P_1\). When a capacitor with capacitance C such that \(X_L = X_C\) is put in series, the power factor becomes \(P_2\). Calculate \(P_1 / P_2\).
If the effective current in a 50 cycle AC circuit is 5 A, what is the peak value of current? What is the current \(\dfrac{1}{600}\) sec after it was zero?
A light bulb is rated 100 W for a 220 V AC supply of 50 Hz. Calculate (a) the resistance of the bulb, (b) the rms current through the bulb.
A 15.0 \(\mu\)F capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what will happen to the capacitive reactance and the current?
An AC circuit consists of only an inductor of inductance 2 H. If the current is represented by a sine wave of amplitude 0.25 A and frequency 60 Hz, calculate the effective potential difference across the inductor. \((\pi = 3.142)\)
An alternating emf of \(e = 220 \sin 100\pi t\) is applied to a circuit containing an inductance of \(\left(\dfrac{1}{\pi}\right)\) henry. Write an equation for the instantaneous current through the circuit. What will be the reading of the AC galvanometer connected in the circuit?
A 25 \(\mu\)F capacitor, a 0.10 H inductor and a 25 \(\Omega\) resistor are connected in series with an AC source whose emf is given by \(e = 310 \sin 314\,t\) (volt). What is the frequency, reactance, impedance, current and phase angle of the circuit?
A capacitor of 100 \(\mu\)F, a coil of resistance 50 \(\Omega\) and an inductance 0.5 H are connected in series with a 110 V, 50 Hz source. Calculate the rms value of current in the circuit.
Find the capacity of a capacitor which, when put in series with a 10 \(\Omega\) resistor, makes the power factor equal to 0.5. Assume an 80 V, 100 Hz AC supply.
Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.
Calculate the value of capacitance in picofarad which will make a 101.4 microhenry inductance oscillate with a frequency of one megahertz.
A 10 \(\mu\)F capacitor is charged to a potential of 25 volt. The battery is disconnected and a pure 100 mH coil is connected across the capacitor so that LC oscillations are set up. Calculate the maximum current in the coil.
A 100 \(\mu\)F capacitor is charged with a 50 V source supply. Then the source supply is removed and the capacitor is connected across an inductance, as a result of which 5 A current flows through the inductance. Calculate the value of the inductance.