## In This post you will get answer of Q.

In a typical Wheatstone’s network the resistance in cyclic order are P = 10 ohm, Q = 5 ohm, S = 4 ohm and R = 4 ohm. For the bridge to balance

Solution:

## and 4 Here $frac{R}{S} = frac{4}{4}$ = 1 and $frac{P}{Q} = frac{10}{5}$

If 5$Omega$ resistor is added in series with Q then $frac{P}{Q} = frac{10}{10} = 1$