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

The Correct Option is A

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, \(I=\frac{V}{R}=\frac{6}{50 / 3}=0.36\, A\)