Four Resistances P,Q,R,X formed a wheat stone bridge. The bridge is balanced when R=100  Ω .If P and Q   are inter changed the bridge balance for121  Ω  .The value of X is

# Four Resistances $\mathrm{P},\mathrm{Q},\mathrm{R},\mathrm{X}$ formed a wheat stone bridge. The bridge is balanced when $\mathrm{R}=100\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{\Omega }$ .If P and Q   are inter changed the bridge balance for$121\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{\Omega }$  .The value of X is

1. A

$100\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Omega }$

2. B

$200\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Omega }$

3. C

$300\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Omega }$

4. D

$110\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Omega }$

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### Solution:

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

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