Question

In Wheatstone bridge, $4$ resistors $P=10\Omega ,Q=5\Omega ,R=4\Omega ,S=4\Omega$ are connected in cyclic order. To ensure no current through galvanometer

• A. $5\Omega$ resistance is connected in series with $Q$
• B. $4\Omega$ resistance is connected parallel to $S$
• C. $10\Omega$ resistance is connected in series with $P$
• D. $4\Omega$ resistance is connected in series with $R$

Answer 1

The correct option is A $5\Omega$ resistance is connected in series with $Q$

For no current through the galvanometer, the wheatstone bridge should be balanced. For this, we must have

$\frac{P}{Q}=\frac{S}{R}$

This condition is satisfied with only option

(a).

When a $5\Omega$ resistor is connected in series with Q, the equivalent resistance in the Q-arm $=5+5=10\Omega$.

$\therefore \frac{P}{Q}=\frac{10}{10}=1$

and $\frac{S}{R}=\frac{4}{4}=1$

$\Rightarrow \frac{P}{Q}=\frac{S}{R}$ which is the requred condition for balanced wheat-stone bridge and thus no current flows through galvanometer.