Question

In the circuit shown the cells $A$ and $B$ have negligible resistances. For $V_A=12V,R_1=500\Omega$ and $R=100\Omega$ the galvanometer $\left(G\right)$ shows no deflection. The value of $V_B$ is :-

• A. $12V$
• B. $6V$
• C. $4V$
• D. $2V$

Answer 1

The correct option is D $2V$

In the first loop which contains $V_A$,the current is given by:

$I=\frac{V_A}{R_1+R}$

$I=\frac{12}{500+100}$

$I=0.02A$

Since negative terminal of battery is considered at zero voltage. Now voltage at the intersection point between $R$ and $R_1$ is given by:

$V=IR$

$V=0.02\times 100$

$V=2V$

So for battery $V_B$ of electric potential $2V$ there will be no current flowing through the galvanometer as voltage difference between these points is zero.