Question

In a meter – bridge circuit as shown, when one more resistance of $100\Omega$ is connected is parallel with unknown resistance ‘x’, then ratio $\frac{{\ell }_1}{{\ell }_2}$ becomes $‘2^{\prime }.{\ell }_1$ is balance length. AB is a uniform wire. Then value of ‘x’ must be

• A. $50\Omega$
• B. $100\Omega$
• C. $200\Omega$
• D. $400\Omega$

Answer 1

The correct option is B

$100\Omega$

$\because$ Wheat stone bridge is in balanced condition

So $\frac{100}{{\ell }_1}=\frac{\frac{100x}{100+x}}{{\ell }_2}\because \frac{{\ell }_2}{{\ell }_1}=2\Rightarrow x=100\Omega$