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)

0.36 A

Answer 2

The correct answer is A:$0.36A$
The balanced condition for Wheatstone's 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
$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$
$\therefore R=\frac{25 \times 50}{25+50}=\frac{50}{3} \,\Omega$
Hence, $I=\frac{V}{R}=\frac{6}{50 / 3}=0.36\, A$