Question

A galvanometer gives full scale deflection of $1\text{volt}$ when acting like a voltmeter when connected in series with $2\text{k}$$\Omega$ resistance. The same galvanometer gives $500\text{mA}$ full scale deflection when acting like a ammeter when connected with shunt resistance of value $0.2\Omega$ in parallel. Find out the resistance of galvanometer (in $\Omega$).

• A. $108$
• B. $222$
• C. $250$
• D. $1.5$

Answer 1

The correct option is B $222$

Let internal resistance of galvanometer is $G$. Galvanometer is connected with $R=2\text{k}\Omega$ in series combination
So. $V=(R+G)I_g$
$\Rightarrow I_g=\frac{V}{(R+G)}$
here, $V=1\text{volt}$ so,$I_g=\frac{1}{(2000+G)}$.......$\left(1\right)$
Now, an ammeter is connected in parallel with a resistance of $0.2\Omega$
So, $Ig\times G=(500\text{mA}-I_g)\times 0.2\Omega$
$\Rightarrow I_g\times G=(0.5A-I_g)\times 0.2$
$\Rightarrow (0.2+G)\times Ig=0.1$
$\Rightarrow (0.2+G)\times \frac{1}{(2000+G)}=0.1$.....[from eq. $\left(1\right)$]
$\Rightarrow (0.2+G)=200+0.1G$
$\Rightarrow 0.9G=199.8\Rightarrow G=222\Omega$
Hence, internal resistance of galvanometer is $222\Omega$