Question

In the circuit shown, $P\ne R$, the reading of the galvanometer is same with switch S open or closed. Then,

• A. $I_R=I_G$
• B. $I_P=I_G$
• C. $I_Q=I_G$
• D. $I_Q=I_R$

Answer 1

The correct option is A $I_R=I_G$

$\text{Step 1:}$

The reading of Galvanometer $G$ is the same with switch $S$ open or closed.

This implies that closing or opening the switch does not effect the circuit. It means that $\text{No current flows through switch}$ $S$.

Hence, the given arrangement of resistors is a $\text{Balanced Wheatstone bridge}$. Hence we can remove switch S.

$\text{Step 2:}$

$R$ and $G$ will be in series.

Hence current flows through resistor $R$ is equal to the current flows through the galvanometer $G$.

$\Rightarrow I_R=I_G$

Hence option (A) is correct.