Question

In the Wheatstone's network given, $ P=10\,\Omega , Q=20\,\Omega ,\,R=15\,\Omega , S=30\,\Omega , $ the current passing through the battery (of negligible internal resistance) is

• A. 0.36 A
• B. zero
• C. 0.18 A
• D. 0.72 A

Answer 1

Answer: (A)

The balanced condition for Wheatstones bridge is
$ \frac{P}{Q}=\frac{R}{S} $
as is obvious from the given values.
No, current flows through galvanometer is zero.
Now, $P$ and $R$ are in series, so
Resistance, $ {{R}_{1}}=P+R $
$ =10+15=25\,\Omega $
Similarly, $Q$ and $S$ are in series, so Resistance
$ {{R}_{2}}=R+S $
$ =20+30=50\,\Omega $
Net resistance of the network as $ {{R}_{1}} $ and $ {{R}_{2}} $ are in parallel
$ i=\frac{V}{R}=\frac{0.6}{50}=0.36\,A $