Physics | VITEEE 2016 | Discussion Page

In Wheatstone bridge, 4 resistors P=10Ω, Q= 5Ω, R=4Ω, S=4Ω are connected in cyclic order.To ensure no current through galvanometer

A) $5\Omega$ resisttance is connected in series Q

B) $4\Omega$ resistance is connected parallel to S

C) $10\Omega$ resistance is connected in series with P

D) $4\Omega$ resistance is connected in series with R

Option A

For no current through the galvanometer, the Wheatstone bridge should be balanced

for this , we must have

$=\frac{P}{Q}=\frac{S}{R}$

This condition is satisfied with only option (a)

when a 5 Ω resistor is connected in series

with Q, the equivalent resistance in the P-arm becomes 10 Ω

$\therefore$ $\frac{P}{Q}=\frac{10}{10}=1$

and $\frac{S}{R}=\frac{4}{4}=1$

$\Rightarrow$ $\frac{P}{Q}=\frac{S}{R}$

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