Question

A galvanometer gives full scale deflection with $0.006\text{A}$ current. By connecting it to a $4990\Omega$ resistance, it can be converted into a voltmeter of range ($0-30\text{V}$). If connected to a $\frac{2X}{249}\Omega$ resistance, it becomes an ammeter of range ($0-1.5\text{A}$). The value of $X$ is ______.

Answer 1

Let $G$ be the resistance of the galvanometer.

To convert galvanometer into voltmeter of range $0-30\text{V}$, a resistance $R=4990\Omega$ is connected in series.


Voltmeter will give maximum reading, when galvanometer is at full deflection.

$V=I_G(G+R)$

$\Rightarrow 30=0.006(G+4990)$

$\Rightarrow G=10\Omega$

To convert the given galvanometer into an ammeter of range $0-1.5\text{A}$, a resistance of value $S=\frac{2X}{249}\Omega$ is connected in parallel as shown in the figure.

Ammeter will give maximum reading, when galvanometer is at full deflection.

$(I-I_G)S=I_G\left(G\right)$

$\Rightarrow (1.5-0.006)\frac{2X}{249}=0.006\left(10\right)$

$\Rightarrow \frac{2X}{249}=\frac{10}{249}$

$\Rightarrow X=5$