Question

The resistance of a galvanometer is $50\Omega$ and the maximum current which can be passed through it is $0.002\text{A}.$ What resistance must be connected to it in order to convert it into an ammeter of range $0-0.5\text{A}$?

• A. $0.002\Omega$
• B. $0.02\Omega$
• C. $0.2\Omega$
• D. $0.5\Omega$

Answer 1

The correct option is C $0.2\Omega$

As we know, $(I-I_G)S=i_{G}G$
Here, $G=50\Omega ;I=0.5\text{A};I_G=0.002\text{A}$
$\Rightarrow S=\frac{0.1}{0.5-0.002}=\frac{100}{498}\simeq 0.2\Omega$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.