VITEEE 2012 Paper

Balanced wheatstone bridge condition

P

Q

=

R

S

No, current flows through galvanometer Now, P and R are in series, so Resistance R1 = P + R
= 10 + 15 = 25 Ω
Similarly, Q and S are in series, so Resistance R2 = R + S = 20 + 30 = 50Ω
Net resistance of the network as R, and R2 are in parallel

1

R

=

1

R1

+

1

R2

∴ R =

25×50

25+50

=

50

3

Ω
Hence, current, I =

V

R

=

6

50∕3

= 0.36 A