Question

A galvanometer of $50-ohm$ resistance has $25$ divisions. A current of$4\times {10}_{-4}$ ampere gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 volts, it should be connected with a resistance of:-

• A. $245\Omega$ as a shunt
• B. $2550\Omega$ in series
• C. $2450\Omega$ in series
• D. $2500\Omega$ as a shunt

Answer 1

The correct option is C $2450\Omega$ in series

Resistance of galvanometer is $R_g=50\Omega$

Total current through galvanometer $I_g=25\times 4\times {10}^{-4}={10}^{-2}A$,

Range of voltage ,$V=25$volt.

For converting galvanometer to voltmeter, we connect a high resistance (HR) in series.

thus, $V=(HR+R_g)I_g$

or $HR=\frac{V}{I_g}-R_g=\frac{25}{{10}^{-2}}-50=2500-50=2450\Omega$