Chapter 9: Current Electricity
In the year 1845, a young physics student at the University of Königsberg sat down with a problem that was quietly troubling engineers all over Europe. Telegraph lines, laboratory circuits and early electrical machines were becoming tangled webs of wires and cells. Such a web could not always be broken down into neat series and parallel groups, and once that was impossible, nobody could say how much current flowed in which wire.
The student was Gustav Robert Kirchhoff, then only about twenty-one years old. Instead of inventing new machinery, he wrote down two simple statements about what a circuit is not allowed to do. Charge cannot collect at a junction, so whatever flows in must flow out. And energy given to a unit charge by the cells must be completely used up as it travels once around any closed loop. From these two ordinary-sounding statements, the current in every branch of even the most complicated network can be calculated.
Remarkably, Kirchhoff wrote these rules about fifty years before the electron was discovered. Yet they are still the rules built into the circuit-simulation software used today to design microchips and to manage national power grids.
Once currents can be predicted, they can also be measured with great precision. That is the second half of this chapter.
Diagram
Fig. 9.0 – A tangled multi-loop circuit of cells and resistors on the left, with two arrows leading to a clean junction diagram and a single closed loop on the right.
Each of these questions is answered somewhere in this chapter. Keep them in mind as you read, and check your answers against the Curiosity Corner boxes that appear along the way.
9.1 Introduction
In Std. XI you studied the origin of electrical conductivity, in particular for metals. You also learnt how to calculate the effective resistance of two or more resistances connected in series and in parallel. In a series combination the same current passes through every resistor, and the effective resistance is the sum of the individual resistances. In a parallel combination the same potential difference appears across every resistor, and the reciprocals of the resistances add up.
These two rules carry you a long way, but only as long as the network can be untangled step by step — one pair of resistors reduced to a single resistor, then the next pair, and so on until a single loop is left.
Where the Series–Parallel Rules Break Down
A circuit containing several complex connections of electrical components cannot always be reduced into a single loop by using the rules of series and parallel combination of resistors. Consider the bridge-type arrangement shown in Fig. 9.1a: notice that no two resistors in it are joined purely end to end, and no two share both their ends.
Diagram
Fig. 9.1a – A five-resistor bridge network across a cell, with the fifth resistor bridging the midpoints of the two branches, marked “cannot be reduced by series–parallel rules”.
The moment the middle resistor is added, the series rule and the parallel rule both fail, because neither condition — same current, or same potential difference — is satisfied by any pair.
More complex circuits can be analysed by using Kirchhoff’s laws. Gustav Robert Kirchhoff (1824–1887) formulated two rules for analysing a complicated circuit. In this chapter we will discuss these laws and their applications.
These laws are also the starting point for several precise measuring arrangements, and in examinations they usually appear either as a multi-loop circuit whose branch currents must be found, or as the basis of a numerical on the Wheatstone bridge, the metre bridge, or the potentiometer.
9.2 Kirchhoff’s Laws of Electrical Network
Before describing these laws we will define some terms used for electrical circuits. Each of these words has an exact meaning here, and applying the laws correctly depends on identifying these parts of a circuit without confusion.
A junction is a meeting point of conductors — a place where the current arriving can divide, or where currents arriving by different paths can combine.
A loop is a path that closes on itself. Starting from any point, you may travel through cells, resistors or plain connecting wire, and if you return to the same point you have traced a loop.
A branch is the stretch of circuit between one junction and the next. Since there is no side path in between for charge to escape into, every element lying on a branch carries the same current.
Reading a Network
These three terms can be picked out directly from a circuit diagram, such as the network drawn in Fig. 9.1, in which two cells of emf \(\varepsilon_1\) and \(\varepsilon_2\) drive current through three resistors of resistance \(R\).
Diagram
Fig. 9.1 – Electric network with cells ε₁ and ε₂, resistors R in the upper and lower paths, a resistor and cell in the middle path, junctions a and b, and the three paths marked 1, 2 and 3.
In Fig. 9.1, there are two junctions, labelled a and b. There are three branches: these are the three possible paths 1, 2 and 3 from a to b. Any two of these paths taken together form a closed conducting path, so the same network also contains loops.
For a steady current flowing through an electrical network of resistors, the following Kirchhoff’s laws are applicable. The condition of a steady current is important: it means the branch currents have settled to constant values and are no longer changing with time.
9.2.1 Kirchhoff’s First Law: (Current law/Junction law)
Charge cannot pile up at a point in a wire. If three amperes of charge arrive at a junction every second, then three amperes must leave it every second, distributed in some way among the other conductors meeting there. Kirchhoff’s first law is simply this statement written in the language of currents.
The algebraic sum of the currents at a junction in an electrical network is zero, i.e.,
\[\sum_{i=1}^{n} I_i = 0\]
where \(I_i\) is the current in the \(i^{\text{th}}\) conductor at a junction having \(n\) conductors. Each current \(I_i\) is measured in ampere (A), and \(n\) is simply the number of conductors meeting at that junction.
The word algebraic is the key. The currents are not all added as plain numbers; each is given a sign depending on whether it moves towards the junction or away from it. Without signs, the sum of several currents could never be zero.
Sign Convention
The currents arriving at the junction are considered positive and the currents leaving the junction are considered negative.
This is a matter of convention, not of physics. The opposite choice would work equally well, provided it is used consistently everywhere in the same problem. What must never happen is a change of convention halfway through a solution.
Applying the Law at a Junction
Consider a junction P in a circuit where six conductors meet, as shown in Fig. 9.2. Notice in Fig. 9.2 that the arrowheads of \(I_1\), \(I_3\) and \(I_4\) point towards P, while those of \(I_2\), \(I_5\) and \(I_6\) point away from it.
Diagram
Fig. 9.2 – Junction P with six conductors; arrows for I₁, I₃, I₄ pointing towards P and arrows for I₂, I₅, I₆ pointing away from P.
Applying the sign convention, we can write
\[I_1 - I_2 + I_3 + I_4 - I_5 - I_6 = 0 \qquad \text{--- (9.1)}\]
Arriving currents \(I_1\), \(I_3\) and \(I_4\) are considered positive and leaving currents \(I_2\), \(I_5\) and \(I_6\) are considered negative.
Equation (9.1) can also be written as
\[I_1 + I_3 + I_4 = I_2 + I_5 + I_6\]
Thus the total current flowing towards the junction is equal to the total current flowing away from the junction. This second form is often the more convenient one in numerical work, because it removes the need to track minus signs.
In examinations, this law rarely appears alone. It is normally used first, to express several unknown currents in terms of fewer variables, before the voltage law is applied.
9.2.2 Kirchhoff’s Voltage Law
Imagine carrying a unit positive charge once around a closed loop of a circuit and bringing it back to the point from which it started. The cells in the loop hand energy to it, and the resistors take energy away from it. Since the charge has returned to the same point, it must be back at the same potential, so the energy gained and the energy lost must balance exactly.
The algebraic sum of the potential differences (products of current and resistance) and the electromotive forces (emfs) in a closed loop is zero.
\[\sum IR + \sum \varepsilon = 0 \qquad \text{--- (9.2)}\]
Here \(I\) is the current through a resistor in ampere (A), \(R\) is its resistance in ohm (\(\Omega\)), so each product \(IR\) is a potential difference in volt (V), and \(\varepsilon\) is the emf of a source in the loop, also in volt (V).
Once again the sum is algebraic. Whether a particular \(IR\) term or emf enters the sum with a plus sign or a minus sign depends on the direction in which we choose to trace the loop, so a sign convention is essential.
Sign Convention
While tracing a loop through a resistor, if we are travelling along the direction of conventional current, the potential difference across that resistance is considered negative. If the loop is traced against the direction of the conventional current, the potential difference across that resistor is considered positive.
The emf of an electrical source is positive while tracing the loop within the source from the negative terminal of the source to its positive terminal. It is taken as negative while tracing within the source from positive terminal to the negative terminal.
The first rule reflects a simple physical fact: conventional current inside a resistor always flows from the higher potential end to the lower potential end. So if you walk with the current, you are walking downhill in potential, and the change is negative. Walk against the current and you climb, so the change is positive.
The second rule works the same way. Inside a cell, the positive terminal is at the higher potential. Entering at the negative terminal and leaving at the positive terminal is a climb, hence positive.
Applying the Voltage Law to a Network
Consider an electrical network shown in Fig. 9.3. Two cells of emf \(\varepsilon_1\) and \(\varepsilon_2\) drive currents \(I_1\) and \(I_2\) through the outer branches, while the current \(I_3\) passes through the resistance \(R_5\) in the middle branch BF.
Diagram
Fig. 9.3 – Electrical network ABCDFG: cell ε₁ with resistors R₁ and R₃ in the left loop carrying I₁, cell ε₂ with resistors R₂ and R₄ in the right loop carrying I₂, and resistor R₅ carrying I₃ in the central branch BF.
Consider the loop ABFGA in clockwise sense. Applying the sign conventions and using Eq. (9.2), we get,
\[-I_1R_1 - I_3R_5 - I_1R_3 + \varepsilon_1 = 0\]
\[\therefore \ \varepsilon_1 = I_1R_1 + I_3R_5 + I_1R_3\]
Each of the three \(IR\) terms carries a minus sign because, in this loop, we travel along the direction of the current in each of those resistors. The emf is positive because we pass through the cell from its negative terminal to its positive terminal.
Now consider the loop BFDCB in anticlockwise direction. Applying the sign conventions, we get,
\[-I_2R_2 - I_3R_5 - I_2R_4 + \varepsilon_2 = 0\]
\[\therefore \ \varepsilon_2 = I_2R_2 + I_3R_5 + I_2R_4\]
Read in words, each of these equations says that the emf supplied by a cell is completely spent as potential drops across the resistors of that loop. Nothing is left over, and nothing appears from nowhere.
Steps Usually Followed While Solving a Problem Using Kirchhoff’s Laws
Choose some direction of the currents.
Reduce the number of variables using Kirchhoff’s first law.
Determine the number of independent loops.
Apply voltage law to all the independent loops.
Solve the equations obtained simultaneously.
In case the answer of a current variable is negative, the conventional current is flowing in the direction opposite to that chosen by us.
Step (ii) is what keeps the algebra manageable. If a junction has three branches, do not introduce three unknowns; write the third current as the sum or difference of the other two, and only two unknowns remain. Step (iii) matters as well: a loop that can be built by combining two loops you have already used gives no new information, so it is not independent and must not be counted.
9.3 Wheatstone Bridge
Resistance of a material changes due to several factors such as temperature, strain, humidity, displacement, liquid level, etc. Therefore, measurement of these properties is possible by measuring the resistance. This is a powerful idea: a quantity that has nothing to do with electricity can be measured electrically, provided it leaves its mark on a resistance.
Measurable values of resistance vary from a few milliohms to hundreds of mega ohms. No single technique works well over such an enormous span. Depending upon the resistance range (milliohm to tens of ohm, tens of ohm to hundreds of ohms, hundreds of ohm to mega ohm, etc.), various methods are used for resistance measurement. Wheatstone’s bridge is generally used to measure resistances in the range from tens of ohm to hundreds of ohms.
The Bridge Circuit
Four resistances P, Q, R and S are connected to form a quadrilateral ABCD as shown in Fig. 9.4. A battery of emf \(\varepsilon\) along with a key is connected between the points A and C such that point A is at higher potential with respect to the point C. A galvanometer of internal resistance G is connected between points B and D.
Diagram
Fig. 9.4 – Wheatstone bridge: quadrilateral ABCD with resistances P (A to B), Q (B to C), S (A to D) and R (D to C); galvanometer G between B and D; cell ε with key between A and C.
The galvanometer forms the “bridge” across the middle of the quadrilateral, and everything about this circuit depends on what it reads.
When the key is closed, current \(I\) flows through the circuit. It divides into \(I_1\) and \(I_2\) at point A. \(I_1\) is the current through P and \(I_2\) is the current through S. The current \(I_1\) gets divided at point B. Let \(I_g\) be the current flowing through the galvanometer. The currents flowing through Q and R are \((I_1 - I_g)\) and \((I_2 + I_g)\) respectively.
Notice that these branch currents have already been written using the junction law, which is why only \(I_1\), \(I_2\) and \(I_g\) appear instead of five separate unknowns. From Fig. 9.4,
\[I = I_1 + I_2 \qquad \text{--- (9.3)}\]
Consider the loop ABDA. Applying Kirchhoff’s voltage law in the clockwise sense shown in the loop we get,
\[-I_1P - I_gG + I_2S = 0 \qquad \text{--- (9.4)}\]
Now consider loop BCDB, applying Kirchhoff’s voltage law in the clockwise sense shown in the loop we get,
\[-(I_1 - I_g)Q + (I_2 + I_g)R + I_gG = 0 \qquad \text{--- (9.5)}\]
From these three equations (Eq. (9.3), (9.4), (9.5)) we can find the current flowing through any branch of the circuit.
The Balanced Bridge
A special case occurs when the current passing through the galvanometer is zero. In this case, the bridge is said to be balanced. Condition for the balance is \(I_g = 0\). This condition can be obtained by adjusting the values of P, Q, R and S.
Physically, \(I_g = 0\) means the points B and D are at exactly the same potential, so there is nothing to push current across the bridge. Substituting \(I_g = 0\) in Eq. (9.4) and Eq. (9.5) we get,
\[-I_1P + I_2S = 0 \quad \therefore \ I_1P = I_2S \qquad \text{--- (9.6)}\]
\[-I_1Q + I_2R = 0 \quad \therefore \ I_1Q = I_2R \qquad \text{--- (9.7)}\]
Dividing Eq. (9.6) by Eq. (9.7), we get
\[\frac{P}{Q} = \frac{S}{R} \qquad \text{--- (9.8)}\]
This is the condition for balancing the Wheatstone bridge.
If any three resistances in the bridge are known, the fourth resistance can be determined by using Eq. (9.8).
Wheatstone Bridge in Practice
Figure 9.4 is a basic circuit diagram of Wheatstone bridge; however, in practice the circuit is used in a different manner. In all cases it is used to determine some unknown resistance. A few applications of Wheatstone bridge circuits are discussed in the following article.
9.3.1 Metre Bridge
The Wheatstone bridge of Fig. 9.4 needs four separate resistance boxes, and adjusting them in the correct ratio is tedious. The metre bridge replaces two of those arms by two parts of a single uniform wire, so that the ratio can be varied smoothly just by sliding a contact along the wire.
Construction
Metre bridge (Fig. 9.5) consists of a wire of uniform cross section and one metre in length, stretched on a metre scale which is fixed on a wooden table. The ends of the wire are fixed below two L shaped metallic strips.
Diagram
Fig. 9.5 – Metre bridge: one metre wire AB on a metre scale, two L-shaped strips with a central strip C forming the left and right gaps, unknown resistance X in the left gap, resistance box R in the right gap, galvanometer from C to jockey J, and cell with key and rheostat across AB.
A single metallic strip separates the two L shaped strips, leaving two gaps, the left gap and the right gap. Usually, an unknown resistance X is connected in the left gap and a resistance box is connected in the right gap.
One terminal of a galvanometer is connected to the central strip C, while the other terminal of the galvanometer carries the jockey (J). Temporary contact with the wire AB can be established with the help of the jockey. A cell of emf \(\varepsilon\) along with a key and a rheostat are connected between the points A and B.
Compare this with Fig. 9.4 and the correspondence becomes clear: X and R are two arms of the Wheatstone bridge, the two parts of the wire on either side of the jockey are the other two arms, and the galvanometer sits across the middle exactly as before.
Locating the Balance Point
A suitable resistance R is selected from the resistance box. The jockey is brought in contact with AB at various points on the wire AB and the balance point (null point), D, is obtained. The galvanometer shows no deflection when the jockey is at the balance point.
Let the respective lengths of the wire between A and D, and that between D and B be \(l_X\) and \(l_R\). Then using the conditions for the balance, we get
\[\frac{X}{R} = \frac{R_{AD}}{R_{DB}}\]
where \(R_{AD}\) and \(R_{DB}\) are the resistances of the parts AD and DB of the wire.
If \(l_X\) and \(l_R\) are the lengths of the parts AD and DB of the wire AB, \(\rho\) is the specific resistance of the wire, and A is the area of cross section of wire AB, then
\[R_{AD} = \frac{\rho \, l_X}{A} \qquad R_{DB} = \frac{\rho \, l_R}{A}\]
Here \(\rho\) is measured in ohm metre (\(\Omega\) m), the lengths in metre (m), and A in square metre (m²). Therefore
\[\frac{X}{R} = \frac{R_{AD}}{R_{DB}} = \frac{\rho \, l_X / A}{\rho \, l_R / A}\]
Since the same wire forms both parts, \(\rho\) and A are identical in numerator and denominator and cancel completely. This is the whole reason the wire must be uniform.
\[\therefore \ \frac{X}{R} = \frac{l_X}{l_R}\]
Therefore,
\[X = \frac{l_X}{l_R} \, R \qquad \text{--- (9.9)}\]
Knowing R, \(l_X\) and \(l_R\), the value of the unknown resistance X can be determined.
Since the wire is one metre long, the two lengths always satisfy \(l_X + l_R = 100\) cm, so measuring one of them is enough.
Sources of Error and How to Reduce Them
Applications
- The Wheatstone bridge is used for measuring the values of very low resistance precisely.
- We can also measure quantities such as galvanometer resistance, capacitance, inductance and impedance using a Wheatstone bridge.
Kelvin’s Method to Determine the Resistance of a Galvanometer
Post Office Box
A post office box (PO Box) is a practical form of Wheatstone bridge as shown in Fig. 9.5c.
Diagram
Fig. 9.5c – Post office box: ratio arms P and Q with plug resistances of 10, 100 and 1000 ohm each, arm R with plugs from 1 to 5000 ohm, unknown resistance X as the fourth arm, galvanometer, cell and the two tap keys K₁ and K₂.
It consists of three arms P, Q and R. The resistances in these three arms are adjustable. The two ratio arms P and Q contain resistances of 10 ohm, 100 ohm and 1000 ohm each. The third arm R contains resistances from 1 ohm to 5000 ohm. The unknown resistance X (usually, in the form of a wire) forms the fourth arm of the Wheatstone’s bridge. There are two tap keys \(K_1\) and \(K_2\).
The resistances in the arms P and Q are fixed to a desired ratio. The resistance in the arm R is adjusted so that the galvanometer shows no deflection. Now the bridge is balanced. The unknown resistance is
\[X = \frac{RQ}{P}\]
where P and Q are the fixed resistances in the ratio arms and R is an adjustable known resistance.
Choosing the ratio \(Q/P\) as 1 : 10, 1 : 1 or 10 : 1 lets the same instrument measure resistances over a wide range with the same set of plugs, which is why this arrangement was convenient enough to be standard equipment in telegraph offices.
If L is the length and r is the radius of the wire X, then the specific resistance of the material of the wire is given by
\[\rho = \frac{X \pi r^{2}}{L}\]
9.4 Potentiometer
A voltmeter is a device which is used for measuring potential difference between two points in a circuit. To read a potential difference, however, a voltmeter must be connected across those two points, and in doing so it offers the current a fresh path of its own.
An ideal voltmeter, which does not change the potential difference to be measured, should have infinite resistance so that it does not draw any current. Practically, a voltmeter cannot be designed to have an infinite resistance. Some current, however small, is always diverted through it, and the potential difference it displays is therefore slightly smaller than the one that existed before it was connected.
Potentiometer is one such device which does not draw any current from the circuit. It acts as an ideal voltmeter. It is used for accurate measurement of potential difference.
9.4.1 Potentiometer Principle
A potentiometer consists of a long wire AB of length L and resistance R having uniform cross sectional area A, as shown in Fig. 9.6. A cell of emf \(\varepsilon\) having internal resistance r is connected across AB. This cell is called the auxiliary or driver cell, and its only task is to maintain a steady current in the wire.
Diagram
Fig. 9.6 – Potentiometer: uniform wire AB of length L connected across a cell of emf ε with internal resistance r and a key, with a point C marked on the wire at a distance l from A.
When the circuit is switched on, current I passes through the wire. The cell, its internal resistance and the wire form a single closed loop, so
\[\text{Current through AB}, \ I = \frac{\varepsilon}{R + r}\]
Potential difference across AB is
\[V_{AB} = IR\]
\[V_{AB} = \frac{\varepsilon R}{(R + r)}\]
Here \(\varepsilon\) is in volt (V), R and r are in ohm (\(\Omega\)), and I is in ampere (A).
Therefore, the potential difference per unit length of the wire is,
\[\frac{V_{AB}}{L} = \frac{\varepsilon R}{L(R + r)}\]
As long as \(\varepsilon\) remains constant, \(\dfrac{V_{AB}}{L}\) will remain constant, because R, r and L are all fixed for a given arrangement. \(\dfrac{V_{AB}}{L}\) is known as the potential gradient along AB and is denoted by K. Potential gradient can be defined as potential difference per unit length of wire. Its SI unit is volt per metre (V m⁻¹), though in laboratory work it is often quoted in volt per centimetre.
Potential Difference Along the Wire
Consider a point C on the wire at distance \(l\) from the point A, as shown in Fig. 9.6. The potential difference between A and C is \(V_{AC}\). Since the same current flows through every part of a uniform wire, each centimetre of it contributes the same potential drop. Therefore,
\[V_{AC} = K l \qquad \text{i.e.} \qquad V_{AC} \propto l\]
Thus, the potential difference between two points on the wire is directly proportional to the length of the wire between the two points, provided (i) the wire is of uniform cross section, (ii) the current through the wire is the same and (iii) the temperature of the wire remains constant.
All three conditions are essential. A non-uniform wire would have a different resistance per centimetre in different parts; a changing current would change the drop per centimetre with time; and a rise in temperature would alter the resistance of the wire, and with it the potential gradient. If any one of these fails, lengths on the scale can no longer be read as potential differences.
The potential gradient therefore depends on the emf of the auxiliary cell, on the resistance of the wire and on the resistance of the rest of the driver circuit, and on the length of the wire — a point that is often asked directly in examinations.
Uses of the potentiometer are discussed below.
9.4.2 Uses of Potentiometer
A) To Compare emf of Cells
Method I
A potentiometer circuit is set up by connecting a battery of emf \(\varepsilon\), with a key K and a rheostat, such that point A is at higher potential than point B. This is the driver circuit that sets up a steady potential gradient along the wire.
The cells whose emfs are to be compared are connected with their positive terminals at point A and negative terminals to the extreme terminals of a two-way key \(K_1 K_2\). The central terminal of the two-way key is connected to a galvanometer. The other end of the galvanometer is connected to a jockey (J), as shown in Fig. 9.7.
Diagram
Fig. 9.7 – Emf comparison by connecting cells individually: potentiometer wire AB with driver cell ε, key K and rheostat; cells ε₁ and ε₂ joined through a two-way key K₁K₂ to a galvanometer and jockey J.
Note that both cells are connected with their positive terminals towards A. This is essential: the two sources must oppose each other along the wire, otherwise a null point can never be obtained.
Key K is closed and then, key \(K_1\) is closed and key \(K_2\) is kept open. Therefore, the cell of emf \(\varepsilon_1\) comes into the circuit. The null point is obtained by touching the jockey at various points on the potentiometer wire AB. Let \(l_1\) be the length of the wire between the null point and the point A. \(l_1\) corresponds to emf \(\varepsilon_1\) of the cell. Therefore,
\[\varepsilon_1 = k \, l_1\]
where k is the potential gradient along the potentiometer wire.
Now key \(K_1\) is kept open and key \(K_2\) is closed. The cell of emf \(\varepsilon_2\) now comes in the circuit. Again, the null point is obtained with the help of the jockey. Let \(l_2\) be the length of the wire between the null point and the point A. This length corresponds to the emf \(\varepsilon_2\) of the cell.
\[\therefore \ \varepsilon_2 = k \, l_2\]
From the above two equations we get
\[\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2} \qquad \text{--- (9.10)}\]
The potential gradient k cancels out, which is why the result does not require k to be known at all. Thus, we can compare the emfs of the two cells. If any one of the emfs is known, the other can be determined.
Method II
The emfs of cells can be compared also by another method called the sum and difference method.
When two cells are connected so that the negative terminal of the first cell is connected to the positive terminal of the second cell, as shown in Fig 9.8 (a), the emfs of the two cells are added up and the effective emf of the combination of two cells is \(\varepsilon_1 + \varepsilon_2\). This method of connecting two cells is called the sum method.
Diagram
Fig. 9.8 (a) – Sum method: negative terminal of the first cell joined to the positive terminal of the second, giving an effective emf of ε₁ + ε₂.
When two cells are connected so that their negative terminals are together, or their positive terminals are connected together, as shown in Fig. 9.8 (b), their emfs oppose each other.
Diagram
Fig. 9.8 (b) – Difference method: like terminals of the two cells joined together, giving an effective emf of ε₁ – ε₂.
In this case their emfs oppose each other and the effective emf of the combination of two cells is \(\varepsilon_1 - \varepsilon_2\) (with \(\varepsilon_1 > \varepsilon_2\) assumed). This method of connecting two cells is called the difference method. Remember that this combination of cells is not a parallel combination of cells.
The circuit is connected as shown in Fig. 9.9. When keys \(K_1\) and \(K_3\) are closed, the cells \(\varepsilon_1\) and \(\varepsilon_2\) are in the sum mode. The null point is obtained using the jockey. Let \(l_1\) be the length of the wire between the null point and the point A. This corresponds to the emf \((\varepsilon_1 + \varepsilon_2)\).
Diagram
Fig. 9.9 – Emf comparison, sum and difference method: potentiometer wire AB with driver circuit, cells ε₁ and ε₂ connected through keys K₁, K₂, K₃, K₄ so that either the sum or the difference combination can be selected, with galvanometer and jockey.
\[\therefore \ \varepsilon_1 + \varepsilon_2 = k \, l_1\]
Now the keys \(K_1\) and \(K_3\) are kept open and keys \(K_2\) and \(K_4\) are closed. In this case the two cells are in the difference mode. Again the null point is obtained. Let \(l_2\) be the length of the wire between the null point and the point A. This corresponds to \(\varepsilon_1 - \varepsilon_2\).
\[\therefore \ \varepsilon_1 - \varepsilon_2 = k \, l_2\]
From the above two equations,
\[\frac{\varepsilon_1 + \varepsilon_2}{\varepsilon_1 - \varepsilon_2} = \frac{l_1}{l_2}\]
By componendo and dividendo method, we get,
\[\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1 + l_2}{l_1 - l_2} \qquad \text{--- (9.11)}\]
Componendo and dividendo is a standard algebraic step: if \(\dfrac{a}{b} = \dfrac{c}{d}\), then \(\dfrac{a + b}{a - b} = \dfrac{c + d}{c - d}\). Applying it here converts the awkward ratio of sums and differences into a direct ratio of the two emfs. Thus, the emfs of two cells can be compared.
B) To Find Internal Resistance (r) of a Cell
The experimental set up for this method consists of a potentiometer wire AB connected in series with a cell of emf \(\varepsilon\), the key \(K_1\), and a rheostat, as shown in Fig. 9.10. The terminal A is at higher potential than terminal B. A cell of emf \(\varepsilon_1\), whose internal resistance \(r_1\) is to be determined, is connected to the potentiometer wire through a galvanometer G and the jockey J. A resistance box R is connected across the cell \(\varepsilon_1\) through the key \(K_2\).
Diagram
Fig. 9.10 – Internal resistance of a cell: potentiometer wire AB with driver cell ε, key K₁ and rheostat; test cell ε₁ with resistance box R and key K₂ across it, connected through galvanometer G and jockey J.
The key \(K_1\) is closed and \(K_2\) is open. The circuit now consists of the cell \(\varepsilon\), cell \(\varepsilon_1\), and the potentiometer wire. The null point is then obtained. Let \(l_1\) be the length of the potentiometer wire between the null point and the point A. This length corresponds to emf \(\varepsilon_1\).
\[\therefore \ \varepsilon_1 = k \, l_1\]
where k is the potential gradient of the potentiometer wire, which is constant.
With \(K_2\) open, no current is drawn from the test cell even outside the balance condition, so the first reading gives its emf.
Now both the keys \(K_1\) and \(K_2\) are closed, so that the circuit consists of the cell \(\varepsilon\), the cell \(\varepsilon_1\), the resistance box, the galvanometer and the jockey. Some resistance R is selected from the resistance box and the null point is obtained. The length of the wire \(l_2\) between the null point and point A is measured. This corresponds to the voltage between the null point and point A.
\[\therefore \ V = k \, l_2 \qquad \therefore \ \frac{\varepsilon_1}{V} = \frac{k \, l_1}{k \, l_2} = \frac{l_1}{l_2}\]
Now the cell is supplying a current through R, so V is its terminal potential difference, which is smaller than \(\varepsilon_1\). The difference between the two is exactly the drop across the internal resistance, and that is what makes the measurement possible.
Consider the loop PQSTP.
\[\varepsilon_1 = IR + Ir \quad \text{and} \quad V = IR\]
\[\therefore \ \frac{\varepsilon_1}{V} = \frac{IR + Ir}{IR} = \frac{R + r}{R} = \frac{l_1}{l_2}\]
\[\therefore \ r = R\left(\frac{l_1}{l_2} - 1\right) \qquad \text{--- (9.12)}\]
This equation gives the internal resistance of the cell. Here R is the resistance taken from the resistance box in ohm (\(\Omega\)), and \(l_1\) and \(l_2\) are the two balancing lengths, which may be in centimetre or metre as long as both are in the same unit.
C) Application of Potentiometer
The applications of the potentiometer discussed above are used in the laboratory. Some practical applications of the potentiometer are given below.
1) Voltage Divider: The potentiometer can be used as a voltage divider to continuously change the output voltage of a voltage supply, as shown in Fig. 9.11. A potential V is set up between points A and B of a potentiometer wire. One end of a device is connected to the positive point A and the other end is connected to a slider that can move along the wire AB. The voltage V divides in proportion of the lengths \(l_1\) and \(l_2\) as shown in Fig. 9.11.
Diagram
Fig. 9.11 – Potentiometer as a voltage divider: supply voltage V across the wire AB, a movable slider dividing the wire into lengths l₁ and l₂, with the output taken between A and the slider.
2) Audio Control: Sliding potentiometers are commonly used in modern low-power audio systems as audio control devices. Both sliding potentiometers (faders) and rotary potentiometers (knobs) are regularly used for frequency attenuation, loudness control and for controlling different characteristics of audio signals.
3) Potentiometer as a sensor: If the slider of a potentiometer is connected to the moving part of a machine, it can work as a motion sensor. A small displacement of the moving part causes a change in potential, which is further amplified using an amplifier circuit. The potential difference is calibrated in terms of the displacement of the moving part.
9.4.3 Advantages of a Potentiometer Over a Voltmeter
Merits
Potentiometer is more sensitive than a voltmeter. A very small change in the quantity being measured shifts the balance point by a measurable length along the wire, whereas a voltmeter may not register the change on its scale at all.
A potentiometer can be used to measure a potential difference as well as an emf of a cell. A voltmeter always measures terminal potential difference, and as it draws some current, it cannot be used to measure the emf of a cell.
Measurement of potential difference or emf is very accurate in the case of a potentiometer. A very small potential difference of the order \(10^{-6}\) volt can be measured with it. The least count of a potentiometer is much better compared to that of a voltmeter.
The reason behind all three merits is the same. A potentiometer does not read a deflection on a scale; it reports a length at which the deflection is zero. Lengths can be measured far more finely than pointer positions, and a null reading is free from errors of scale calibration.
Demerits
The potentiometer is not portable, since it needs a long wire mounted on a board along with a driver cell and a rheostat. Direct measurement of potential difference or emf is also not possible with it: the instrument gives a balancing length, and the required value must then be calculated from that length using the potential gradient.
For examination purposes, the comparison between the two instruments is best remembered through this contrast — a voltmeter draws current and gives a direct reading, while a potentiometer draws no current at balance and gives an indirect but far more accurate result.
9.5 Galvanometer
Every measurement described so far in this chapter has depended on one instrument noticing that a current is zero, or very nearly so. That instrument is the galvanometer.
A galvanometer is a device used to detect weak electric currents in a circuit. It has a coil pivoted (or suspended) between the concave pole faces of a strong laminated horse shoe magnet, as shown in Fig. 9.12.
Diagram
Fig. 9.12 – Internal structure of a galvanometer: a coil pivoted between the concave pole faces of a horse shoe magnet, with a soft iron core, a spring and a pointer moving over a scale.
When an electric current passes through the coil, it deflects. The current-carrying coil sits in the magnetic field of the permanent magnet, and the magnetic force on it produces a torque that turns the coil against a restoring spring. The coil settles where the two effects balance.
The deflection is proportional to the current passing through the coil. This proportionality is what makes the instrument useful for measurement and not merely for detection, because equal steps of current then produce equal steps of deflection.
The deflection of the coil can be read with the help of a pointer attached to it. The position of the pointer on the scale provided indicates the current passing through the galvanometer or the potential difference across it. Thus, a galvanometer can be used as an ammeter or a voltmeter with suitable modification.
Why Modification Is Needed
The galvanometer coil has a moderate resistance (about 100 ohms) and the galvanometer itself has a small current carrying capacity (about 1 mA).
Both of these figures are limitations. A current of a few amperes would burn out a coil built to carry a milliampere, and a resistance of about 100 ohms is far too large for an instrument meant to be inserted into a circuit and far too small for one meant to be connected across it.
The next two articles show how the same instrument, with one resistance added in the right place, becomes either an ammeter reading amperes or a voltmeter reading volts.
9.5.1 Galvanometer as an Ammeter
Let the full scale deflection current and the resistance of the coil G of a moving coil galvanometer (MCG) be \(I_g\) and G. Full scale deflection current means the current that swings the pointer right across to the last mark of the scale; any larger current cannot be shown, and may damage the coil.
It can be converted into an ammeter, which is a current measuring instrument. An ammeter is always connected in series with a resistance R through which the current is to be measured, because only then does the whole of that current pass through the instrument.
To Convert a Moving Coil Galvanometer (MCG) into an Ammeter
To convert an MCG into an ammeter, the modifications necessary are
Its effective current capacity must be increased to the desired higher value.
Its effective resistance must be decreased. The finite resistance G of the galvanometer, when connected in series, decreases the current through the resistance R which is actually to be measured. In the ideal case, an ammeter should have zero resistance.
Care must be taken to protect it from possible damage due to the passage of an excessive electric current.
All three requirements point in the same direction: most of the current must be kept out of the coil, and the instrument as a whole must offer almost no obstruction to the circuit.
In practice this is achieved by connecting a low resistance in parallel with the galvanometer, which effectively reduces the resistance of the galvanometer. This low resistance connected in parallel is called shunt (S). This arrangement is shown in Fig. 9.13.
Diagram
Fig. 9.13 – Ammeter: galvanometer of resistance G with a low shunt resistance S connected in parallel across it, total current I entering, I_g through the galvanometer and I_s through the shunt.
Uses of the Shunt
It is used to divert a large part of the total current by providing an alternate path, and thus it protects the instrument from damage.
It increases the range of an ammeter.
It decreases the resistance between the points to which it is connected.
Calculating the Shunt Resistance
The shunt resistance is calculated as follows. In the arrangement shown in Fig. 9.13, \(I_g\) is the current through the galvanometer. Therefore, the current through S is
\[I_s = (I - I_g)\]
Since S and G are in parallel, the potential difference across them is the same,
\[G I_g = S I_s\]
\[\therefore \ G I_g = S (I - I_g)\]
\[\therefore \ S = \left(\frac{I_g}{I - I_g}\right) G \qquad \text{--- (9.13)}\]
Here I is the total current to be measured in ampere (A), \(I_g\) is the full scale deflection current of the galvanometer in ampere (A), and G and S are in ohm (\(\Omega\)).
Equation 9.13 is useful to calculate the range of current that the galvanometer can measure.
- If the current I is n times the current \(I_g\), then \(I = n I_g\). Using this in the above expression we get
\[S = \frac{G I_g}{n I_g - I_g} \quad \text{or,} \quad S = \frac{G}{n - 1}\]
This is the required shunt to increase the range n times. Notice that the larger the range required, the smaller the shunt must be — an ammeter of wide range carries a very small shunt resistance indeed.
- Also if \(I_s\) is the current through the shunt resistance, then the remaining current \((I - I_s)\) will flow through the galvanometer. Hence
\[G (I - I_s) = S I_s\]
i.e.
\[G I - G I_s = S I_s\]
i.e.
\[S I_s + G I_s = G I\]
\[\therefore \ \frac{I_s}{I} = \frac{G}{S + G}\]
This equation gives the fraction of the total current through the shunt resistance.
9.5.2 Galvanometer as a Voltmeter
A voltmeter is an instrument used to measure the potential difference between two points in an electrical circuit. It is always connected in parallel with the component across which the voltage drop is to be measured, because two elements in parallel share the same potential difference. A galvanometer can be used for this purpose.
To Convert a Moving Coil Galvanometer into a Voltmeter
To convert an MCG into a voltmeter, the modifications necessary are:
Its voltage measuring capacity must be increased to the desired higher value.
Its effective resistance must be increased, and
It must be protected from possible damage, which is likely due to excess applied potential difference.
All these requirements can be fulfilled if we connect a resistance of suitably high value (X) in series with the given MCG. The series resistance takes up most of the applied potential difference, leaving only a small share across the coil, and at the same time limits the current through it.
A voltmeter is connected across the points where the potential difference is to be measured. If a galvanometer is used to measure voltage, it draws some current (due to its low resistance), therefore the actual potential difference to be measured decreases. To avoid this, a voltmeter should have very high resistance. Ideally, it should have infinite resistance.
Diagram
Fig. 9.14 – Voltmeter: galvanometer of resistance G with a high resistance X connected in series with it, the combination joined across the two points whose potential difference V is to be measured.
A very high resistance X is connected in series with the galvanometer for this purpose as shown in Fig. 9.14. The value of the resistance X can be calculated as follows.
If V is the voltage to be measured, then
\[V = I_g X + I_g G\]
The applied potential difference is shared between the added resistance and the coil, and the same current \(I_g\) passes through both because they are in series.
\[\therefore \ I_g X = V - I_g G\]
\[\therefore \ X = \frac{V}{I_g} - G \qquad \text{--- (9.14)}\]
where \(I_g\) is the current flowing through the galvanometer, in ampere (A), V is in volt (V), and G and X are in ohm (\(\Omega\)). Eq. (9.14) gives the value of resistance X.
If \(n_V = \dfrac{V}{V_g} = \dfrac{V}{I_g G}\) (where \(V > V_g\)) is the factor by which the voltage range is increased, it can be shown that
\[X = G (n_V - 1)\]
Comparison of an Ammeter and a Voltmeter
| AMMETER | VOLTMETER | |
|---|---|---|
| 1. | It measures current. | It measures potential difference. |
| 2. | It is connected in series. | It is connected in parallel. |
| 3. | It is an MCG with low resistance. (Ideally zero) | It is an MCG with high resistance. (Ideally infinite) |
| 4. | Smaller the shunt, greater will be the current measured. | Larger its resistance, greater will be the potential difference measured. |
| 5. | Resistance of ammeter is \(R_A = \dfrac{S \cdot G}{S + G} = \dfrac{G}{n}\) | Resistance of voltmeter is \(R_V = G + X = G \cdot n_V\) |
9.6 Summary
Circuit Terms
- A junction is any point in an electric circuit where two or more conductors are joined together. A loop (or mesh) is any closed conducting path in an electric network. A branch is any part of the network lying between two junctions; every element of a branch carries the same current.
Kirchhoff’s Laws
- These laws apply to a steady current flowing through an electrical network of resistors.
- First law (junction law): the algebraic sum of the currents at a junction is zero, \(\sum_{i=1}^{n} I_i = 0\). Currents arriving at the junction are taken positive and those leaving are taken negative. Equivalently, the total current flowing towards a junction equals the total current flowing away from it.
- Second law (voltage law): the algebraic sum of the potential differences and the emfs in a closed loop is zero, \(\sum IR + \sum \varepsilon = 0\).
- Sign convention for the voltage law: travelling through a resistor along the direction of conventional current, the potential difference is negative; against the current, it is positive. An emf is positive when the loop is traced within the source from the negative terminal to the positive terminal, and negative when traced from positive to negative.
- The first law expresses conservation of electric charge; the second law expresses conservation of energy.
- A negative value for a current means the conventional current flows opposite to the direction assumed.
Wheatstone Bridge
- Four resistances P, Q, R and S form a quadrilateral ABCD, with a cell between A and C and a galvanometer of resistance G between B and D.
- \(I = I_1 + I_2\); loop ABDA gives \(-I_1P - I_gG + I_2S = 0\); loop BCDB gives \(-(I_1 - I_g)Q + (I_2 + I_g)R + I_gG = 0\).
- The bridge is balanced when \(I_g = 0\). Then \(I_1P = I_2S\) and \(I_1Q = I_2R\), giving the balance condition
\[\frac{P}{Q} = \frac{S}{R}\]
- If any three resistances are known, the fourth can be found. The balance condition does not involve the emf of the cell or the galvanometer resistance.
- The bridge is generally used for resistances from tens of ohms to hundreds of ohms, and also for calibrating voltmeters and ammeters, for measuring very low resistances precisely, and for measuring galvanometer resistance, capacitance, inductance and impedance.
Metre Bridge
- A metre bridge is a practical Wheatstone bridge in which two arms are the two parts of a uniform wire one metre long. With the unknown resistance X in the left gap and a known resistance R in the right gap,
\[\frac{X}{R} = \frac{R_{AD}}{R_{DB}} = \frac{l_X}{l_R} \qquad \therefore \ X = \frac{l_X}{l_R}\,R\]
where \(R_{AD} = \dfrac{\rho l_X}{A}\) and \(R_{DB} = \dfrac{\rho l_R}{A}\), and \(l_X + l_R = 100\) cm.
- Sources of error: non-uniform cross section of the wire, contact resistance at the soldered ends, and inaccurate measurement of \(l_X\) and \(l_R\). Errors are minimised by keeping the null point in the middle one third of the wire (between 34 cm and 66 cm), by repeating the experiment with X and R interchanged, and by tapping rather than sliding the jockey.
- Kelvin’s method determines a galvanometer resistance using a metre bridge:
\[G = \left(\frac{l_g}{100 - l_g}\right) R \qquad \{l_g + l_r = 100 \ \text{cm}\}\]
- Post office box: a practical Wheatstone bridge with adjustable ratio arms P and Q (10, 100, 1000 ohm each) and arm R (1 to 5000 ohm), giving \(X = \dfrac{RQ}{P}\), and for a wire of length L and radius r, \(\rho = \dfrac{X \pi r^{2}}{L}\).
Potentiometer
- A potentiometer draws no current from the circuit at balance and therefore acts as an ideal voltmeter, unlike a real voltmeter which has finite resistance.
- For a wire AB of length L and resistance R driven by a cell of emf \(\varepsilon\) and internal resistance r,
\[I = \frac{\varepsilon}{R + r}, \qquad V_{AB} = \frac{\varepsilon R}{(R + r)}, \qquad \frac{V_{AB}}{L} = \frac{\varepsilon R}{L(R + r)}\]
- Potential gradient \(K = \dfrac{V_{AB}}{L}\) is the potential difference per unit length, in V m⁻¹. For a point C at a distance \(l\) from A, \(V_{AC} = K l\), so \(V_{AC} \propto l\).
- This proportionality holds provided (i) the wire is of uniform cross section, (ii) the current through the wire is the same, and (iii) the temperature of the wire remains constant.
- Comparing emfs, individual method: \(\varepsilon_1 = k l_1\) and \(\varepsilon_2 = k l_2\), hence
\[\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2}\]
- Comparing emfs, sum and difference method: with \(\varepsilon_1 + \varepsilon_2 = k l_1\) and \(\varepsilon_1 - \varepsilon_2 = k l_2\),
\[\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1 + l_2}{l_1 - l_2}\]
- Internal resistance of a cell: with \(l_1\) the balancing length for the open cell and \(l_2\) that with external resistance R connected,
\[r = R\left(\frac{l_1}{l_2} - 1\right)\]
- Practical uses: voltage divider, audio control (faders and knobs), and motion sensing.
- Merits over a voltmeter: greater sensitivity, ability to measure emf as well as potential difference, and high accuracy — potential differences of the order of \(10^{-6}\) V can be measured. Demerits: it is not portable and does not give a direct reading.
Galvanometer and Its Conversion
- A galvanometer detects weak currents; its coil is pivoted between the concave pole faces of a strong laminated horse shoe magnet and its deflection is proportional to the current. The coil has a moderate resistance (about 100 ohms) and a small current capacity (about 1 mA).
- As an ammeter: a low resistance shunt S is connected in parallel. Since \(G I_g = S(I - I_g)\),
\[S = \left(\frac{I_g}{I - I_g}\right) G\]
- For a range increased n times (\(I = n I_g\)), \(S = \dfrac{G}{n - 1}\), and the fraction of current through the shunt is \(\dfrac{I_s}{I} = \dfrac{G}{S + G}\).
- The shunt protects the instrument, increases the range of the ammeter, and decreases the resistance between the points across which it is connected.
- As a voltmeter: a high resistance X is connected in series. Since \(V = I_g X + I_g G\),
\[X = \frac{V}{I_g} - G\]
- For a voltage range increased \(n_V\) times, \(X = G(n_V - 1)\).
- Resistance of the ammeter \(R_A = \dfrac{S \cdot G}{S + G} = \dfrac{G}{n}\); resistance of the voltmeter \(R_V = G + X = G \cdot n_V\). An ammeter is connected in series and should ideally have zero resistance; a voltmeter is connected in parallel and should ideally have infinite resistance.
Thermoelectricity
- Seebeck discovered that two dissimilar metals joined to form a closed loop, with their junctions at different temperatures, produce a small thermo emf. Such a pair is called a thermocouple, and the effect is reversible.
- With the cold junction at \(0^{\circ}\)C and the hot junction at \(T^{\circ}\)C,
\[\varepsilon = \alpha T + \frac{1}{2}\beta T^{2}\]
where \(\alpha\) and \(\beta\) are the thermoelectric constants, so the graph of \(\varepsilon\) against temperature is a parabola.
9.7 Objective Questions
Kirchhoff’s first law, i.e., \(\Sigma I = 0\) at a junction, deals with the conservation of (A) charge (B) energy (C) momentum (D) mass
When the balance point is obtained in the potentiometer, a current is drawn from (A) both the cells and auxiliary battery (B) cell only (C) auxiliary battery only (D) neither cell nor auxiliary battery
In the following circuit diagram, an infinite series of resistances is shown. Equivalent resistance between points A and B is (A) infinite (B) zero (C) 2 Ω (D) 1.5 Ω
Diagram
Fig. 9.E1 – Infinite ladder network: a repeating chain of 1 Ω resistors along the top line from A, with 2 Ω resistors connected vertically down to the bottom line from B, both lines continuing to infinity.
Four resistances 10 Ω, 10 Ω, 10 Ω and 15 Ω form a Wheatstone’s network. What shunt is required across the 15 Ω resistor to balance the bridge (A) 10 Ω (B) 15 Ω (C) 20 Ω (D) 30 Ω
A circular loop has a resistance of 40 Ω. Two points P and Q of the loop, which are one quarter of the circumference apart, are connected to a 24 V battery having an internal resistance of 0.5 Ω. What is the current flowing through the battery? (A) 0.5 A (B) 1 A (C) 2 A (D) 3 A
To find the resistance of a gold bangle, two diametrically opposite points of the bangle are connected to the two terminals of the left gap of a metre bridge. A resistance of 4 Ω is introduced in the right gap. What is the resistance of the bangle if the null point is at 20 cm from the left end? (A) 2 Ω (B) 4 Ω (C) 8 Ω (D) 16 Ω
9.8 Check Your Concepts
Why is the sum in Kirchhoff’s first law called an algebraic sum? What would go wrong if all the currents meeting at a junction were simply added as positive numbers?
A student applies the voltage law to a loop by tracing it anticlockwise, while another traces the same loop clockwise. Show, by reasoning, why both obtain the same branch currents.
On solving a network, a student obtains one of the branch currents as −0.8 A. What does the negative sign tell you? Is the magnitude of the current wrong?
Explain why the balancing condition of a Wheatstone bridge contains neither the emf of the cell nor the resistance of the galvanometer, and why this is an advantage.
What is meant by a null method of measurement? State two reasons why a null method gives a more accurate value of resistance than a method based on reading an ammeter and a voltmeter.
In deriving the metre bridge formula, the specific resistance \(\rho\) and the area of cross section A of the wire cancel out. Which physical assumption about the wire makes this cancellation valid, and what happens to the result if that assumption fails?
In a metre bridge experiment the null point is obtained at 8 cm from the left end. Should the resistance in the box be increased or decreased for a more reliable reading? Justify your answer.
Why is a metre bridge experiment repeated after interchanging the positions of the unknown resistance and the resistance box? Which source of error does this address?
In Kelvin’s method, why is the balance point identified by “no change in deflection” rather than by “zero deflection”? What does this condition tell you about the two points concerned?
State and explain the effect on the potential gradient of a potentiometer wire when (i) the emf of the auxiliary cell is increased, (ii) an extra resistance is included in series with the potentiometer wire, and (iii) a longer wire of the same material and cross section is used for the same driver circuit.
While comparing the emfs of two cells with a potentiometer, both cells are connected with their positive terminals joined to the higher-potential end of the wire. What would be observed if one of the cells were connected the other way round, and why?
A potentiometer is described as an ideal voltmeter. Justify this description by considering the current drawn from the cell at the balance point.
In the potentiometer experiment for internal resistance, the emf of the test cell is given in the question but is never used in the calculation. Explain why it is not required.
Why does a longer potentiometer wire, or a smaller potential gradient, allow a small potential difference to be measured more accurately?
Explain why an ammeter must be connected in series and must have a very low resistance, while a voltmeter must be connected in parallel and must have a very high resistance. What error arises in each case if this requirement is not met?
As the required range of an ammeter is made larger, the shunt resistance needed becomes smaller. Explain this using the relation between the shunt and the range factor n.
Assertion: Kirchhoff’s junction law is a statement of conservation of electric charge. Reason: In the steady state, charge cannot accumulate at any junction of a network. Examine whether the reason correctly explains the assertion.
A student repeatedly slides the jockey along the metre bridge wire instead of tapping it. Explain how this practice affects the results obtained in later experiments performed with the same wire.
In a thermocouple, what happens to the direction of the current if the hot and cold junctions are interchanged? What property of the thermoelectric effect does this illustrate?
Two identical galvanometers are converted, one into an ammeter and the other into a voltmeter. Compare the resistances of the two finished instruments, and state which of them may safely be connected directly across a cell.
Define or describe a potentiometer.
Define potential gradient.
Why should the jockey not be slided along the potentiometer wire?
Are Kirchhoff’s laws applicable for both AC and DC currents?
In a Wheatstone’s metre-bridge experiment, the null point is obtained in the middle one third portion of the wire. Why is it recommended?
State any two sources of errors in a metre-bridge experiment. Explain how they can be minimized.
What is potential gradient? How is it measured? Explain.
On what factors does the potential gradient of the wire depend?
Why is a potentiometer preferred over a voltmeter for measuring emf?
State the uses of a potentiometer.
What are the disadvantages of a potentiometer?
Distinguish between a potentiometer and a voltmeter.
What will be the effect on the position of zero deflection if only the current flowing through the potentiometer wire is (i) increased (ii) decreased?
Obtain the balancing condition in case of a Wheatstone’s network.
Explain with a neat circuit diagram how you will determine the unknown resistance by using a metre-bridge.
Describe Kelvin’s method to determine the resistance of a galvanometer by using a metre bridge.
Describe how a potentiometer is used to compare the emfs of two cells by connecting the cells individually.
Describe how a potentiometer is used to compare the emfs of two cells by the combination method.
Describe with the help of a neat circuit diagram how you will determine the internal resistance of a cell by using a potentiometer. Derive the necessary formula.
On what factors does the internal resistance of a cell depend?
9.9 Practice with Numericals
Five conductors meet at a junction. Currents of 8 A, 3 A and 2 A flow into the junction, and a current of 6 A flows out through the fourth conductor. Find the magnitude and direction of the current in the fifth conductor.
A cell of emf 12 V and negligible internal resistance is connected in a single loop with resistors of 3 Ω, 4 Ω and 5 Ω joined one after another. Using the loop law, find the current in the circuit and the potential difference across each resistor, and verify that these potential differences add up to the emf.
Two cells of emf 6 V and 4 V, having internal resistances 1 Ω and 2 Ω respectively, are connected in parallel with their like poles joined together. The combination sends current through an external resistance of 3 Ω. Using Kirchhoff’s laws, find the current through each cell and through the external resistance.
In a two-loop network, a 10 V cell of negligible internal resistance drives current through a 5 Ω resistor in the left branch and a 20 V cell of negligible internal resistance drives current through a 10 Ω resistor in the right branch, the two branches meeting at a junction from which a 4 Ω resistor leads back to the common point. Find the current in each of the three branches.
In a Wheatstone bridge ABCD, P = 20 Ω, Q = 30 Ω and S = 45 Ω. Find the value of R for which the galvanometer connected between B and D shows no deflection.
A Wheatstone bridge has ratio arms of 100 Ω and 10 Ω, and the adjustable arm reads 250 Ω at balance. Find the unknown resistance forming the fourth arm.
In a metre bridge, a resistance of 5 Ω is placed in the right gap and the null point is found at 60 cm from the left end. Find the unknown resistance in the left gap. Where would the null point lie if the two gap resistances were interchanged?
In a metre bridge experiment, the balancing length measured from the left end is 35 cm when a resistance of 4 Ω is used in the right gap. Calculate the unknown resistance. If an error of 1 mm is made in locating the null point, estimate the percentage error in the length measurement.
A metre bridge wire is 1 m long and has a diameter of 0.08 cm. Its total resistance is 2 Ω. Calculate the specific resistance of the material of the wire. (Take π = 3.142)
In Kelvin’s method, a known resistance of 30 Ω is used in one gap of a metre bridge and the galvanometer in the other. The balance point is obtained at 40 cm from the end opposite the resistance box, so that the length of wire opposite the galvanometer is 40 cm. Calculate the resistance of the galvanometer.
In a post office box, the ratio arms are set at P = 100 Ω and Q = 10 Ω, and the bridge balances when R = 4530 Ω. Find the unknown resistance X. If X is a wire of length 1.2 m and radius 0.2 mm, calculate the specific resistance of its material.
A potentiometer wire is 8 m long and has a resistance of 16 Ω. It is connected in series with a cell of emf 4 V and internal resistance 4 Ω. Calculate the current through the wire, the potential difference across it, and the potential gradient in V/m.
Two cells are compared by the individual method on a potentiometer. The balancing lengths obtained are 120 cm and 150 cm. If the emf of the first cell is 1.2 V, find the emf of the second cell.
Two cells connected in the sum mode on a potentiometer give a balancing length of 300 cm, and in the difference mode the balancing length is 100 cm. Compare the emfs of the two cells.
In an experiment to find the internal resistance of a cell, the balancing length with the external circuit open is 90 cm. When a resistance of 5 Ω is connected across the cell, the balancing length becomes 75 cm. Calculate the internal resistance of the cell.
A galvanometer of resistance 80 Ω gives a full scale deflection for a current of 2 mA. Find the shunt resistance required to convert it into an ammeter reading up to 2 A, and calculate the fraction of the total current that then passes through the shunt.
A galvanometer of resistance 100 Ω shows full scale deflection for a current of 1 mA. Find the resistance to be connected in series with it to convert it into a voltmeter of range 0 to 15 V, and find the factor by which the voltage range has been increased.
A galvanometer has a resistance of 25 Ω and gives a full scale deflection for a current of 10 mA. Find the resistance required, and state how it must be connected, to convert the galvanometer (a) into an ammeter of range 0 to 5 A and (b) into a voltmeter of range 0 to 25 V.
A galvanometer of resistance 60 Ω is shunted by a resistance of 0.5 Ω. Calculate the effective resistance of the ammeter so formed, and the factor by which the current range of the galvanometer has been increased.
The potential gradient along a potentiometer wire is 2 mV per centimetre. The emf of a cell is balanced at a length of 45 cm. Find the emf of the cell. If the same cell, when shunted by a 4 Ω resistor, balances at 36 cm, find its internal resistance.
A battery of emf 4 volt and internal resistance 1 Ω is connected in parallel with another battery of emf 1 V and internal resistance 1 Ω (with their like poles connected together). The combination is used to send current through an external resistance of 2 Ω. Calculate the current through the external resistance.
Two cells of emf 1.5 volt and 2 volt, having respective internal resistances of 1 Ω and 2 Ω, are connected in parallel so as to send current in the same direction through an external resistance of 5 Ω. Find the current through the external resistance.
A voltmeter has a resistance of 30 Ω. What will be its reading when it is connected across a cell of emf 2 V having internal resistance 10 Ω?
A set of three coils having resistances 10 Ω, 12 Ω and 15 Ω are connected in parallel. This combination is connected in series with a series combination of three coils of the same resistances. Calculate the total resistance and the current through the circuit, if a battery of emf 4.1 volt is used for drawing current.
A potentiometer wire has a length of 1.5 m and resistance of 10 Ω. It is connected in series with a cell of emf 4 volt and internal resistance 5 Ω. Calculate the potential drop per centimetre of the wire.
When two cells of emfs \(\varepsilon_1\) and \(\varepsilon_2\) are connected in series so as to assist each other, their balancing length on a potentiometer is found to be 2.7 m. When the cells are connected in series so as to oppose each other, the balancing length is found to be 0.3 m. Compare the emfs of the two cells.
The emf of a cell is balanced by a length of 120 cm of potentiometer wire. When the cell is shunted by a resistance of 10 Ω, the balancing length is reduced by 20 cm. Find the internal resistance of the cell.
The potential drop per unit length along a wire is \(5 \times 10^{-3}\) V/m. If the emf of a cell balances against a length of 216 cm of this potentiometer wire, find the emf of the cell.
The resistance of a potentiometer wire is 8 Ω and its length is 8 m. A resistance box and a 2 V battery are connected in series with it. What should be the resistance in the box, if it is desired to have a potential drop of 1 μV/mm?
Find the equivalent resistance between the terminals F and B in the network shown in the figure below, given that the resistance of each resistor is 10 ohm.
Diagram
Fig. 9.E2 – Resistor network with lettered terminals including F and B, every resistor of the network being 10 Ω.
- A voltmeter has a resistance of 100 Ω. What will be its reading when it is connected across a cell of emf 2 V and internal resistance 20 Ω?