Potentiometer: Principle, Applications, Formula & Solved Examples (Complete Guide)

20.8 POTENTIOMETER

20.8.1 Potensiometer

Figure 20.49 A uniform wire AB with several ohms of resistance is connected in series with a constant DC source (acid accumulator), a rheostat, and a switch. This arrangement allows a steady current to flow through the wire, producing a uniform potential gradient along AB for measurement purposes.

1. Circuit

A simple potentiometer may consist of a uniform bare wire AB with several ohms of resistance. The wire is connected in series to a constant dc supply such as an acid accumulator, a rheostat and a switch, as shown in Figure 20.49.

2. Principle

(a) A constant current flows through the potentiometer wire AB, thus producing a p.d. V_AB across AB.

(b) Consider any two points, D and E, on the wire which are separated by a distance l. A p.d. V_DE exists across D and E, which is given by

V_DE = (R_DE / R_AB) V_AB

We have

R_DE = ρ(DE) / A

where ρ and A are the resistivity of the material of wire AB and the cross-sectional area of the wire respectively.

R_AB = ρ(AB) / A

R_DE / R_AB = DE / AB

Hence,

V_DE = (DE / AB) V_AB

= (V_AB / AB) DE

If the current in the whole circuit and the temperature of wire AB are constant, V_AB and AB will remain constant all the time. Hence, we get

V_DE ∝ l (DE = l)

In other words, the p.d. across any two points on the potentiometer wire AB is directly proportional to the distance between the two points.

In general, the p.d. V across any two points on the potentiometer wire separated by distance l is given by

V = kl

where k is a constant.

EXAMPLE 20.14

Refer to the circuit shown in Figure 20.49. The potentiometer wire AB is 100 cm long and the p.d. across it is 1.5 V. Determine the distance between two points on the wire across which a p.d. of 0.75 V is produced.

Answer

V = kl

V_AB = k(AB)

V / V_AB = l / AB

0.75 / 1.5 = l / 100

l = 50 cm

20.8.2 Applications of Potentiometer – Comparison of emf

Figure 20.50 Two electric cells $C_1$ and $C_2$ with emf $E_1$ and $E_2$ respectively are connected to a potentiometer through a switch. The arrangement allows each cell to be connected alternately so that their emf can be compared by measuring the corresponding balancing lengths on the potentiometer wire

1 Circuit

Two electric cells C₁ and C₂ of emf E₁ and E₂ respectively are connected to a potentiometer in the manner as shown in Figure 20.50.

2 Method

(a) The switch is allowed to make contact with point X so that only C₁ is connected to the potentiometer.

(b) The jockey is allowed to slide on the potentiometer wire AB until a balanced point D where the galvanometer indicates zero deflection is located. The length AD (= l₁) is measured.

(c) The switch is then allowed to make contact with point Y so that now only C₂ is connected to the potentiometer.

(d) The balanced point for this cell is located. The length l₂ between A and the balanced point is measured.

3 Principle

When the jockey is at the balanced point D the galvanometer indicates zero deflection. This means that no current flows through both the galvanometer and the cell C₁. Hence, we have

emf of C₁ = p.d. across AD

E₁ = kl₁

E₂ = kl₂

E₁ / E₂ = l₁ / l₂

Note: If emf, say, E₂ is known, then we can determine emf E₁.

EXAMPLE 20.15

An electric cell of emf 2.0 V and internal resistance 0.50 Ω is used to send current to the potentiometer wire. The wire has length 100 cm and resistance 4.0 Ω. Determine the length of the wire which can provide the correct p.d. to balance the emf of an electric cell. The emf of the cell is 1.08 V.

Answer

When balance is achieved, let the length of the potentiometer wire be l and the p.d. across this length be V. Then we have

V = kl

where k is a constant. When the potentiometer is under balanced condition, we have

emf of cell = V

1.08 = kl .......... (1)

Let the p.d. across the whole length AB of the wire be V_AB. Then we have

V_AB = k(AB)

= k(100 cm) .... (2)

(1) ÷ (2),

1.08 / V_AB = l / 100

l = 108 / V_AB

Using the potential divider equation, we get

$V_{AB}=\left(\frac{R_{AB}}{R_{AB}+r}\right)E=\frac{4}{4+0.5}(0.2)=1.78$

$l=\frac{108}{1.78}=60.7 \ cm$

20.8.3 Applications of Potentiometer – Measuring the Internal Resistance of an Electric Cell

Figure 20.51 An electric cell C with emf E and internal resistance r is connected to the potentiometer along with a resistance box and a switch. This setup allows the determination of the internal resistance of the cell by comparing the balancing lengths under open-circuit and loaded conditions.

1 Circuit

Figure 20.51 shows how the electric cell C of emf E and internal resistance r is connected to the potentiometer.

2 Method

(a) Switch S is kept open and the jockey connected to the galvanometer G is moved along the potentiometer wire AB. The balanced point D is located when G indicates zero reading. The length of AD (= l₀) is measured.

(b) Switch S is then closed so that a resistance R provided by a resistance box is connected parallel to C. The new location of the balanced point is determined. The distance l of the balanced point from A is measured.

(c) The procedure above is repeated using several different values of R and the corresponding values of l are measured.

(d) A graph of 1/R is plotted against l. The internal resistance r of the cell C can be determined from the graph.

3 Principle

(a) Suppose switch S is open. When balance is achieved, no current flows through the galvanometer G and the cell. This means that the emf E of cell C is exactly equal to the p.d. V_AD across AD. Hence, we have

E = V_AD

= kl₀ .......... (1)

where k is a constant.

(b) Next, suppose switch S is closed. Now a current I flows through the cell and so a p.d. V is developed across resistance R. When balance is achieved, we get

V = V_AD
= kℓ .......................... (2)

(1) ÷ (2)

Applying the potential divider equation to R and the cell C, we get

V = ( R / (R + r) ) E

E / V = (R + r) / R = I₀ / I

1 + (r/R) = I₀ / I

1/R = (1/r) ( I₀ / I − 1 )

= ( I₀ / r )(1/I) − 1/r

This equation is equivalent to the linear equation:

y = mx − c

if we let y = 1/R, x = 1/l, m = I₀/r and c = 1/r.

Figure 20.52 A graph of 1/R plotted against 1/l produces a straight slanting line. The gradient of the line represents $I_0/r$ while the intercept on the 1/R axis represents 1/r, allowing the internal resistance of the cell to be determined experimentally.

If we plot a graph of 1/R against 1/l, we will get a slanting line, as shown in Figure 20.52.

The line has the following features:

(a) gradient of line = I₀ / r

(b) intercept on the 1/R axis = 1 / r

Hence, the internal resistance r of the cell can be determined experimentally using the graph.

EXAMPLE 20.16

A 2-ohm resistor is connected directly across the terminals of an electric cell. When the p.d. across the cell is applied to the sliding wire of a potentiometer, a balanced point is obtained at a distance of 80 cm from one end of the potentiometer. Another 2-ohm resistor is then connected directly across the first resistor. It is found that now the balanced point is located at distance 65 cm from the same end of the potentiometer. Determine the internal resistance of the cell.

Figure 20.53 An electric cell with emf E and internal resistance r connected to a potentiometer with a single 2 Ω resistor across it.

Answer

Figure 20.53 shows how the circuit is to be connected. The cell has emf E and internal resistance r.

Only one 2-ohm resistor connected across cell:

When balance is achieved, we have

p.d. across cell = p.d. across balanced length

V₁ = k (80 cm)

where k is a constant. Applying the potential divider equation to the 2 ohm resistor, we get

$V_{1}=\left(\frac{R_{1}}{R_{1}+r}\right)E$

$=\left(\frac{2}{2+r}\right)E$

$\left(\frac{2}{2+r}\right)E=k(80 \ cm)$ . . . . (1)

With two 2-ohm resistors connected across cell:

When balance is achieved, we get

$V_2=k(65 \ cm)$

But $V_{2}=\left(\frac{R_{2}}{R_{2}+r}\right)E$

and $\frac{1}{R_2}=\frac{1}{2}+\frac{1}{2}=1$

$R_2=1 \ \Omega$

$\left(\frac{1}{1+r}\right)E=k(65 \ cm)$ . . . . (2)

(1) : (2),

$\frac{2(1+r)}{2+r}=\frac{80}{65}$

Solving, we get r = 0.6 Ω.

20.8.4   Application of Potentiometer – Comparison of Low Resistances

Figure 20.54 Two low resistors R₁ and R₂ are connected in series with an acid accumulator and a rheostat, then linked to a potentiometer to compare their resistances.

1 Circuit

Two resistors of low resistances R₁ and R₂ are connected in series to each other. They are connected to an acid accumulator S and a rheostat, as shown in Figure 20.54. This circuit is then connected to a potentiometer.

2 Method

(a) The end X of resistor R₁ is connected to the end A of the potentiometer wire AB. The end Y is connected to a galvanometer G, which in turn is connected to a jockey.

(b) The jockey is moved on the wire AB until the galvanometer indicates zero deflection at a point D. The rheostat is adjusted until the length AD is greater than half the length of the wire AB. The distance l₁ of D from A is measured.

(c) The connection between X and A is removed. The end Y is now connected to A and the end Z is connected to G.

(d) The jockey is again moved on wire AB and a new balanced point is obtained. The distance l₂ of the balanced point from A is measured.

3 Principle

X connected to A:
Suppose a constant current I flow through both the resistors. When the circuit is balanced we have

p.d. across R₁ = p.d. across AD

IR₁ = k l₁ ............ (1)

Y connected to A:

When the circuit is balanced, we have

p.d. across R₂ = p.d. across AD

IR₂ = kl₂ ............ (2)

(1) ÷ (2),

R₁ / R₂ = l₁ / l₂

20.8.5   Application of Potentiometer – Measuring Very Low emf

1 Adapting the potentiometer

We need to adapt the potentiometer which has a bare sliding wire of 100 cm in order to use it to measure voltages whose values are only several millivolts. Such low voltages are produced by a thermocouple.

Reason:
Suppose a voltage of 2.0 V is applied across the 100 cm wire of the potentiometer. We have

2.0 V = k(100 cm)

where k is a constant. Then a voltage of, say, 10 mV is developed across a length l, given by

0.010 V = kl

or,

l = (0.010 V / 2.0 V)(100 cm) = 0.5 cm

Such short length is highly undesirable because large errors are incurred in measuring so small a length of the sliding wire. If possible we would like to produce a p.d. of 10 mV across 50 cm or more than 50 cm of the sliding wire.

2 Circuit

Figure 20.55 The potentiometer with a 100 cm sliding wire XY is modified by adding a standard cell and a resistance box to enable accurate measurement of very small emf values.

The circuit is shown in Figure 20.55. XY is the 100 cm sliding wire of the potentiometer. Two components which have been added to the potentiometer are

(i) the standard cell of emf E_s
(ii) the resistance box R

3 Method

(a) The resistance of the whole length of the sliding wire XY is determined. The resistance per cm r of the wire is then calculated.

(b) Suppose that we intend to measure an emf of several millivolt. Then a suitable value of R is chosen so that a p.d. of about 10 mV is developed across XY.

(c) Switch S is allowed to make contact with point B. The jockey J is allowed to move on the sliding wire until the galvanometer G indicates zero deflection. The distance XJ (= l₁) of the balanced point from X is measured.

(d) Switch S is then allowed to make contact with point A. The new balanced point is located and the distance l₂ of the point from X is measured.

4 Principle

(a) Suppose that an acid accumulator of emf 2.0 V and negligible internal resistance is connected across points W and Y.

(b) We intend to let the current I flow through the sliding wire to produce a p.d. of 10 mV across XY. This means that the resistance R has to be such a value that a p.d. of 1.99 V can be developed across R.

(c) Let switch S make contact with point B. We are letting the emf of the standard cell to oppose the p.d. across WJ. When balance is achieved, we have

emf of standard cell = p.d. across WX + p.d. across XJ

= IR + I(resistance of XJ)

E_s = I(R + r₁) .......... (1)

(d) Let switch S make contact with point A. We are now letting the emf E of the thermocouple to oppose the p.d. across XJ. When balance is achieved, we have

emf of thermocouple = p.d. across XJ

= I(resistance of XJ)

E = I(r₂) .......... (2)

(2) ÷ (1),

E / E_s = r₂ / (R + r₁)

The emf E of the thermocouple can be determined since the values of E_s, r, R, l₁ and l₂ are known.

EXAMPLE 20.17

Refer to the circuit shown in Figure 20.55. An acid accumulator of emf 2.00 V and negligible internal resistance is connected across WY. The potentiometer wire XY has length 100 cm and resistance 0.040 ohm per cm. A p.d. of about 10 mV is to be developed across XY. What must be the value of R?

Answer

When balance is achieved, we have

p.d. across R + p.d. across XY = emf of accumulator

p.d. across R = 2.00 − 0.01 = 1.99

Applying the potential divider equation to R, we get

p.d. across R = ( R / (R + rl) ) E

( R / (R + rl) ) (2) = 1.99

2R = 1.99[R + (0.040)(100)]

= 1.99R + 8

R = 8 / 0.01 = 800 Ω

EXAMPLE 20.18

A potentiometer is used to measure the emf of a thermocouple. A constant current flows through the potentiometer whose sliding wire has a resistance of 4.00 Ω m⁻¹. A coil of resistance 1 000 Ω is connected in series to the sliding wire. Balance is achieved when an emf of 1.018 V provided by a Weston cell is applied across the coil together with 450 cm of the sliding wire. Balance is also achieved when the thermocouple is connected only across 120.5 cm of the sliding wire. Determine

(a) the current that flows through the sliding wire

(b) the emf of the thermocouple.

Answer

(a) Figure 20.56 shows how the circuit is to be set up.

Figure 20.56 The circuit shows a potentiometer setup where a known emf of 1.018 V is applied across a coil and a portion of the sliding wire. The arrangement is used to establish a balance condition for measurement.

Consider the emf of 1.018 V being applied across the coil together with 450 cm of the sliding wire. When balance is achieved, we have

emf of 1.018 V = p.d. across coil + p.d. across sliding wire

$1.018 = IR + I(rl_1)$

$=I(1 \ 000)+I(4.00)(4.50)=1 \ 018I$

$I=\frac{1.018}{1 \ 018}=1.00 \times 10^{-3} \ A$

(b) Consider the emf V of the thermocouple applied across the wire. When balance is achieved, we have

V = p.d. across 120.5 cm  of wire

$= I(rl_2)

$=(1.00 \times 10^{-3})(4.0)(1.025)=4.82 \times 10^{-3} \ V$

Applications of Potentiometer

1. Comparison of emf

A potentiometer can be used to compare the electromotive force (emf) of two cells without drawing current from them.

• Two cells with emf E₁ and E₂ are connected alternately.
• The balancing lengths l₁ and l₂ are measured.

Principle:

E₁ = kl₁

E₂ = kl₂

Thus,

E₁ / E₂ = l₁ / l₂

This method is very accurate because no current is drawn from the cells during measurement.

2. Measuring Internal Resistance of a Cell

The potentiometer is used to determine the internal resistance of an electric cell.

• First, measure balancing length l₀ with switch open (no current).
• Then measure balancing length l when a resistor R is connected.

Principle:

E = kl₀

V = kl

Using the relation:

V = ( R / (R + r) ) E

The internal resistance r can be determined experimentally, often using a graph of 1/R against 1/l.

3. Comparison of Low Resistances

The potentiometer can compare two low resistances R₁ and R₂ connected in series.

• Measure balancing lengths l₁ and l₂.

Principle:

IR₁ = kl₁

IR₂ = kl₂

Thus,

R₁ / R₂ = l₁ / l₂

4. Measuring Very Low emf (e.g., Thermocouple)

Potentiometers can be adapted to measure very small voltages such as millivolt outputs from thermocouples.

• A standard cell and resistance box are used to control current.
• Two balancing lengths l₁ and l₂ are measured.

Principle:

E_s = I(R + r₁)

E = I(r₂)

Thus,

E / E_s = r₂ / (R + r₁)

This allows precise determination of very small emf values.

5. General Advantage of Potentiometer

• Highly accurate measurement of voltage.
• No current drawn from the source during balance condition.
• Suitable for measuring very small emf.
• Used in calibration of voltmeters and ammeters.

Conclusion

The potentiometer is a highly precise instrument used to measure potential difference and electromotive force (emf) based on the principle that the potential difference along a uniform wire is directly proportional to its length.

Unlike ordinary measuring devices, the potentiometer does not draw current from the source during measurement, ensuring high accuracy and reliability.

It has wide applications such as comparing emf of cells, determining internal resistance, comparing low resistances, and measuring very small voltages like those produced by thermocouples.

Overall, the potentiometer is an essential tool in electrical and physics experiments due to its sensitivity, accuracy, and versatility.

Frequently Asked Questions (FAQ) Potensiometer 

1. What is a potentiometer?

A potentiometer is an electrical instrument used to measure potential difference (voltage) or electromotive force (emf) by balancing it against a known voltage without drawing current from the source.

2. What is the working principle of a potentiometer?

The potentiometer works on the principle that the potential difference across a uniform wire is directly proportional to its length, provided the current remains constant.

3. Why is a potentiometer more accurate than a voltmeter?

A potentiometer is more accurate because it measures voltage under a null condition (no current flows through the circuit being measured), thus eliminating errors due to internal resistance.

4. What are the main applications of a potentiometer?

It is used to compare emf of cells, measure internal resistance, compare low resistances, and measure very small voltages such as those from thermocouples.

5. What is meant by balance point?

The balance point is the position on the potentiometer wire where the galvanometer shows zero deflection, indicating that the unknown emf is equal to the potential difference across that length of wire.

6. What happens if the current in the potentiometer wire changes?

If the current changes, the potential gradient (k) also changes, which affects the accuracy of measurements. Therefore, a constant current is required.

7. Why is a long wire used in a potentiometer?

A longer wire allows for more precise measurement because it increases the length corresponding to a given voltage, reducing percentage error.

8. Can a potentiometer measure very small voltages?

Yes, with proper adjustments such as adding resistance and using a standard cell, a potentiometer can measure very small voltages in the millivolt range.

9. What is the role of a rheostat in a potentiometer circuit?

The rheostat is used to control and maintain a constant current in the potentiometer wire.

10. Why does the galvanometer show zero at balance point?

At the balance point, the potential difference across the potentiometer wire exactly equals the emf being measured, so no current flows through the galvanometer, resulting in zero deflection.

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