Question

Four resistances of $3\Omega ,3\Omega ,3\Omega$ and $4\Omega$ respectively are used to form a Wheatstone bridge. The $4\Omega$ resistance is short-circuited with a resistance R in order to get bridge balance. The value of R will be.

• A. $10\Omega$
• B. $11\Omega$
• C. $12\Omega$
• D. $13\Omega$

Answer 1

The correct option is A $10\Omega$

Four resistance, $R_1=R_2=R_3=3\Omega$

$R_4=4\Omega$

Resistance R is connecting with $4\Omega$

$R_{eq}=\frac{4R}{4+R}$ ........$\left(1\right)$

for balance bridge

$\frac{R_1}{R_2}=\frac{R_{eq}}{R_3}$

$\frac{3}{3}=\frac{R_{eq}}{3}$

$R_{eq}=3$

From equation $\left(1\right)$

$R_{eq}=3=\frac{4R}{4+R}$

$12+3R=4R$

$R=12\Omega$.