When a resistance of $5 ~\Omega$ is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of $250 \mathrm{~mA}$, however when $1050 ~\Omega$ resistance is connected with it in series, it gives full scale deflection for 25 volt. The resistance of galvanometer is ____________ $\Omega$.
Answer (integer)
50
Solution
Given:
<br/><br/>
$\frac{250 \ \text{mA} \times 5}{5 + R_G} = i$
<br/><br/>
$i = \frac{25}{1050 + R_G}$
<br/><br/>
Equating the two expressions for current, $i$:
<br/><br/>
$\frac{250 \ \text{mA} \times 5}{5 + R_G} = \frac{25}{1050 + R_G}$
<br/><br/>
This equation simplifies to:
<br/><br/>
$100(5 + R_G) = 1050 \times 5 + R_G \times 5$
<br/><br/>
Solving for the resistance of the galvanometer, $R_G$:
<br/><br/>
$95 R_G = 4750$
<br/><br/>
$R_G = 50 \ \Omega$
<br/><br/>
So, the resistance of the galvanometer is $50 \ \Omega$.
About this question
Subject: Physics · Chapter: Current Electricity · Topic: EMF and Internal Resistance
This question is part of PrepWiser's free JEE Main question bank. 120 more solved questions on Current Electricity are available — start with the harder ones if your accuracy is >70%.