Formulas/physics/Moving Charges/Series Resistance for Voltmeter Conversion

Series Resistance for Voltmeter Conversion

High resistance R in series gives full-scale deflection at terminal voltage V. Very high total resistance minimises loading on the circuit under measurement.

Class 12

Derivation

Problem

A galvanometer with resistance $G$ gives full-scale deflection at current $I_g$ . We want full-scale deflection when voltage across the voltmeter terminals is $V$ .

Setup

Connect a high resistance $R$ in series with the galvanometer. The full-scale current $I_g$ must flow through the series combination when $V$ is applied across the terminals.

Derivation

By Ohm's law:

V = I_g(R + G)

R + G = \frac{V}{I_g}

\boxed{R = \frac{V}{I_g} - G}

Net Resistance of Voltmeter

R_V = R + G = \frac{V}{I_g}

Since $I_g$ is very small (galvanometer is sensitive), $R_V$ is very large — voltmeter draws negligible current from the circuit.

Extending the Range

To measure voltage $nV$ (where $n > 1$ ) with a voltmeter already calibrated for $V$ , add additional series resistance $(n-1)R_V$ .

Note

An ideal voltmeter has infinite resistance. A real voltmeter should have resistance much larger than the impedance of the circuit being measured. The series resistor achieves this.

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