Question

To find the resistance of a galvanometer by the half deflection method the following circuit is used with resistances $R_1=9970\Omega ,R_2=30\Omega$ and $R_3=0$. The deflection in the galvanometer is $d$. With $R_3=107\Omega$ the deflection changed to $\frac{d}{2}$. The galvanometer resistance is approximately:

• A. $107\Omega$
• B. $137\Omega$
• C. $107/2\Omega$
• D. $77\Omega$

Answer 1

The correct option is D $77\Omega$

$\text{Solution}:$

At initial condition, $R_3=0$

Equivalent resistance of $R_2$ & $R_3$ = 30 $\Omega$

When $R_3=107\Omega$ & $R_2=30\Omega$,

Then equivalent resistance should be $\frac{30}{2}$= $15\Omega$

This will be when equivalent resistance , $R_g,R_3$ will be parallel to $R_2$, we get resistance of $15\Omega$

Let $R_g-R_3=R_{eq}$

$\frac{1}{R_2}+\frac{1}{R_{eq}}=\frac{1}{30}+\frac{1}{30}=\frac{1}{15}$

$R_{eq}=30\Omega$

$\therefore R_3-R_g=30$

$\therefore 107-R_g=30$

$\therefore R_g=77\Omega$

$\text{Hence D is the correct option}$