Discussion Forum

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

Answer:

Explanation:

 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 $\frac{S}{R}=\frac{4}{4}=1$

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