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One of the circuits for the measurement of resistance by potentiometer is shown. The galvanometer is connected at point A and zero deflection is observed at length PJ = 30 cm. In second case the secondary cell is changed.
Take E_S = 10 V and $r = 1 Ω$ in 1st reading and E_S = 5 V and $r = 2 Ω$ in 2nd reading. In second case, the zero deflection is observed at length PJ = 10 cm. What is the resistance R (in ohm)?

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$4 Ω$

$3 Ω$

$1 Ω$

$2 Ω$

answer is A.

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Detailed Solution

When there is no current through galvanometer, current through R is

$I = \frac{E_{S}}{R + r}$

Potential difference $= I R = \frac{E_{S} R}{R + r}$

Now $\frac{E_{S} R}{R + r} = K l$

For first case: $\frac{10 R}{R + 1} = K (30)$ ….(i)

For second case: $\frac{5 R}{R + 2} = K (10)$ …..(ii)

Dividing (i) and (ii), we get R = $1 Ω$

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