Question

In the circuit, the galvanometer $ G $ shows zero deflection. If the batteries $ A $ and $ B $ have negligible internal resistance, then the value of $ R $ is

• A. $ 100 \, \Omega $
• B. $ 200 \, \Omega $
• C. $ 500 \, \Omega $
• D. $ 1000 \, \Omega $

Answer 1

Answer: (A)

According to given figure,
Applying $KVL$ in loop $(i) $
$500\, I_{1}+R(I_{1}-I_{2})=12$
$(500+R) I_{1}-RI_{2}=12 \dots (i)$
Applying $KVL$ in loop (ii), we have
$R(I_{2}-I_{1})=-2$
$RI_{2}=RI_{1}-2 \dots(ii)$
From Eqs. $(i)$ and $(ii)$, we get
$(500+R) I_{1}-(RI_{1}-2)=12$
$500I_{1}+RI_{1}-RI_{1}+2=12$
$500I_{1}=10$
$I_{1}=\frac{1}{50}$
From Eq. (ii),
$RI_{2}=R\times \frac{1}{50}-2$
But given that galvanometer $G$ shows zero deflection, hence $I_{2}=O$
$\therefore 0=\frac{R}{50}-2$
$\frac{R}{50}=2$
$R=100\,\Omega$