Question

A potentiometer circuit is shown in figure. The jockey of the potentiometer is kept exactly at the mid-point and the ammeter reads the current I, as 0.2 mA. When $V_x$ is reversed, the ammeter reads 3.8 mA. The internal resistance of the ammeter and the unknown potential $V_x$ are respectively

• A. $500\Omega ,1.1V$
• B. $500\Omega ,1.1V$
• C. $500\Omega ,0.9V$
• D. $50\Omega ,0.9V$

Answer 1

The correct option is D $50\Omega ,0.9V$


Writing KVL in outer loop,

$2-I_1\times 900-IR-V_x=0$ ...(i)
(Here R is resistance of ammeter)

In inner loop,

$2-900I_1-900(I_1-I)=0$

$1+450I=990I_1$ ...(ii)

from (i) and (ii) we have

$2-(1+450I)-IR-V_x=0$

$1-450I-IR-V_x=0$ ...(iii)

$I=0.2mA$

So from equation (iii),

$1-450\times 0.2\times {10}^{-3}-0.2\times {10}^{-3}R-V_x=0$

$1000-90-0.2R-1000V_x=0$

$0.2R+1000V_x=910$ ...(iv)

When direction of $V_x$ is reversed then from equation (iii),

$1-450I-IR+V_x=0$ ...(v)

$\text{Here, I = 3.8 mA}$

so from equation (v),

$1-450\times 3.8\times {10}^{-3}-3.8\times {10}^{-3}R+V_x=0$

$1000-1710-3.8R+1000V_x=0$

$-3.8R+1000V_x=710$ ...(vi)

Solving, (iv) and (v) we get,

$R=50\Omega$

$V_x=0.9V$