Q. 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
VITEEEVITEEE 2016
Solution:
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.
