Potentiometer
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Potentiometer
Current Electricity of Class 12
A potentiometer is an instrument that measures the terminal potential difference with high accuracy without drawing any current from the unknown source. It is based on the principle that if constant current is passed through a wire of uniform cross-section, then potential difference across any segment of the wire is proportional to its length.
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The given diagram shows a typical arrangement to measure emf Ex of a battery. The wire ab is of uniform cross−section and carries a constant current supplied by battery S. First the switch K1, is closed and K2 is kept open. The slider is moved on the wire ab till we get zero deflection in the galvanometer. If C1 is the corresponding point in the wire, Ex = Vac1. |
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Now, the experiment is repeated with key K1 open and K2 closed. This time, if the null deflection is obtained with contact on wire at C2,
Eo = Vac2 (Eo is known)
Now,, where lac1 and lac2 are the lengths of segments ac1 and ac2 respectively.
Determination of Internal Resistance of a Cell by Potentiometer Method:
We can also use a potentiometer to measure internal resistance of a cell as shown in the figure. For this the cell (emf ε) whose internal resistance (r) is to be determined is connected across a resistance box through a key K2, as shown in the figure. With key K2 open, balance is obtained at length l1 (AN1). Then,
ε = φ l1…(i)
[where φ is the potential drop per unit length]
when key K2 is closed, the cell sends a current (I) through the resistance box (R.B.). If V is the terminal potential difference of the cell and balance is obtained at length l2(AN2).
V = φl2…(ii)
So, we have ε/V = l1/l2…(iii)
But, ε = I(r + R) and V = IR. This gives [where R is the resistance of the R.B.]
ε/V = (r + R)/R …(iv)
from equation (iii) and (iv) we have
(R + r)/R = l1/l2
r = R[(l1/l2) −1]…(v)
using eq. (v) we can find the internal resistance of a given cell. The potentiometer has the advantage that it draws no current from the voltage source to be measured. As such, it is unaffected by the internal resistance of the source.
Illustration 34.A 10 m long wire of uniform cross – section and 20 Ω resistance is fitted in a potentiometer. The wire is connected in series with battery of 5 volt, alongwith an external resistance of 480 Ω If an unknown emf E is balanced at 6.0 m length of this wire, calculate (i) the potential gradient of the potentiometer wire, (ii) the value of the unknown emf E.
Solution:I = 5/480 + 20A = 0.01A
P.D. across the potentiometer wire = (0.01 × 20 ) V = 0.2 V
(i) Potential gradient = 0.2V/10m = 0.02 Vm−1
(ii) E = (0.02 × 6.0) V = 0.12 volt.
Exercise 11:With a certain cell, the balance point is obtained at 65 cm from the end of a potentiometer wire. With another cell whose emf differs from that of first by 0.1 V, the balanced point is obtained at 60 cm. Find the emf of each cell.
