Question

In a practical meter bridge circuit as shown, when one more resistance of $50\Omega$ is connected parallel with unknown resistance ${}^{\prime }{x}^{\prime }$, then ratio $\frac{l_1}{l_2}$ becomes ${}^{\prime }{2}^{\prime }$. The value of x is $(l_1$ is balance length, AB is uniform wire)

• A. $100\Omega$
• B. $50\Omega$
• C. $200\Omega$
• D. $25\Omega$

Answer 1

The correct option is B $50\Omega$

In balanced Wheatstone bridge,


$\frac{R_1}{R_2}=\frac{R_3}{R_4}$

Here, $R_3=\frac{50x}{50+x}$

$\Rightarrow \frac{50}{l_1}=\frac{\left[\frac{50x}{50+x}\right]}{l_2}$

$\because l_1:l_2=2:1$

$\Rightarrow \frac{50}{l_1}=\frac{\left[\frac{50x}{(50+x)}\right]}{\left(\frac{l_1}{2}\right)}$

$2500+50x=100x$

$x=50\Omega$