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. $0.36A$
• B. Zero
• C. $0.18A$
• D. $0.72A$

Answer 1

Answer: (A)

$0.36A$

Balanced wheatstone bridge condition
$\frac{P}{Q}=\frac{R}{S}$
No, current flows through galvanometer
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, current, $I=\frac{V}{R}=\frac{6}{50/3}=0.36A$