In the Wheatstone's network given, 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, P=10 Ω , Q=20 Ω , R=15 Ω , S=30 Ω , the current passing through the battery (of negligible internal resistance) is (A) 0.36 A (B) zero (C) 0.18 A (D) 0.72 A

Tardigrade Admin · Accepted Answer

The balanced condition for Wheatstones 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 Ω

Similarly,

Q

and

S

are in series, so Resistance

R_{2} = R + S

= 20 + 30 = 50 Ω

Net resistance of the network as

R_{1}

and

R_{2}

are in parallel

i = \frac{V}{R} = \frac{0.6}{50} = 0.36 A

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

0.36 A

zero

0.18 A

0.72 A

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