Four Resistances $\mathrm{P},\mathrm{Q},\mathrm{R},\mathrm{X}$ formed a wheat stone bridge. The bridge is balanced when $\mathrm{R}=100\mathrm{\Omega }$ .If P and Q   are inter changed the bridge balance for$121\mathrm{\Omega }$  .The value of X is

Solution:

Resistance $R=100\mathrm{\Omega }$ Resistance $R=121\mathrm{\Omega }$
Balancing condition for wheat stone bridge
$\frac{P}{Q}=\frac{R}{X}\Rightarrow \frac{P}{Q}=\frac{100}{X}...\left(1\right)$

$$\begin{gathered} P\text{ and }Q\text{ are inter changed }\frac{Q}{P}=\frac{R}{X} \\ \frac{Q}{P}=\frac{121}{X}....\left(2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{on} 1\times 2 \\ \mathrm{we} \mathrm{get} \frac{\mathrm{P}}{\mathrm{Q}}\times \frac{\mathrm{Q}}{\mathrm{P}}=\frac{100}{\mathrm{X}}\times \frac{121}{\mathrm{X}}\Rightarrow 1=\frac{100}{\mathrm{X}}\mathrm{x}\frac{121}{\mathrm{X}}\therefore \mathrm{X}=110\mathrm{\Omega } \end{gathered}$$