A galvanometer has an internal resistance of 20\text{ }\Omega and produces a full-scale de… — JAMB UTME

A galvanometer has an internal resistance of $20\text{ }\Omega$ and produces a full-scale deflection for a current of $15\text{ mA}$ . What resistance of the multiplier, in ohms ( $\Omega$ ), is required to convert this galvanometer into a voltmeter capable of measuring a maximum potential difference of $30\text{ V}$ ?

Answer: 1980 \Omega

Answer

The required multiplier resistance is

1980\text{ }\Omega

To convert a galvanometer into a voltmeter, a multiplier resistor

R_m

is connected in series. The maximum potential difference

V

across the combination is given by

V = I_g(R_g + R_m)

. Substituting

V = 30\text{ V}

I_g = 0.015\text{ A}

, and

R_g = 20\text{ }\Omega

yields

R_m = 1980\text{ }\Omega

Step-by-Step Solution

Convert current from milliamperes to amperes.

I_g = 15\text{ mA} = 0.015\text{ A}

Standard electrical formulas require current in amperes.

Apply the relationship between voltage range, galvanometer current, internal resistance, and multiplier resistance.

V = I_g(R_g + R_m)

, where

V = 30\text{ V}

I_g = 0.015\text{ A}

, and

R_g = 20\text{ }\Omega

A multiplier resistor is connected in series with the galvanometer so that the total potential difference is distributed across both components.

Rearrange the equation and compute

R_m

R_m = \frac{30}{0.015} - 20 = 2000 - 20 = 1980\text{ }\Omega

Subtracting the internal resistance of the galvanometer from the total required resistance yields the necessary multiplier resistance.

Key Concept

Galvanometer conversion to a voltmeter using a series multiplier resistor

Estimated Time:1m 30s

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