Question

A galvanometer coil has a resistance of $50\Omega$ and the meter shows full scale deflection for a current of $5mA$, This galvanometer is converted into voltmeter of range $0-20V$ by connecting

• A. $3950\Omega$ in series with galvanometer
• B. $4050\Omega$ in series with galvanometer
• C. $3950\Omega$ in parallel with galvanometer
• D. $4050\Omega$ in parallel with galvanometer

Answer 1

Answer: (A)

Here,
$R=?$
$G=50\Omega$
$V=20V$
$I_g=5\times 10^{-3}A$
where, $G$ : Resistance of the galvanometer
$V$ : Required range of the voltmeter
$I_g:$ Current in galvanometer which gives full deflection
$R:$ High resistance in series with galvanometer
Applying the formula
$R=\frac{V}{I_g}-G$
$=\left(\frac{20}{5\times 10^{-3}}\right)-50$
$=4000-50$
$=3950\Omega$
This resistance is connected with galvanometer in series.