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

Answer: (A)

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.