VITEEE 2016 Paper

For no current through the galvanometer, the wheatstone bridge should be balanced. For this, we must have

P

Q

=

S

R

⇒

10

5

=

4

4

2 = 1 is not balanced so we need to add 5 to Q then the circuit will be balanced.
This condition is satisfied with only option (a).

When a 5Ω resistor is connected in series
with Q, the equivalent resistance in the P-arm becomes 1 Ω
∴

P

Q

. =

10

10

= 1
and

S

R

=

4

4

= 1
⇒

P

Q

=

S

R