Question

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

• A. 0.72 A
• B. 0 A
• C. 0.18 A
• D. 0.36 A

Answer 1

Answer: (D)

0.36 A

Answer 2

The balanced condition for Wheatstone's bridge is given by $\frac{P}{Q} = \frac{R}{S}$
As is obvious from the given values condition, 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, $I = \frac{V}{R} = \frac{6}{50/3} = 0.36A$