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

Question

In the Wheatstone's network given below P=10 Ω Q=20 Ω R=15 Ω S=30 Ω 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

Tardigrade Admin · Accepted Answer

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 = 25Ω

Similarly, Q and S are in series, so
Resistance

R_{2} = R + S = 20 + 30 = 50Ω

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}}

∴ R = \frac{25 \times 50}{25 + 50} = \frac{50}{3} Ω

Hence,

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

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

0.72 A

0 A

0.18 A

0.36 A

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