A galvanometer has an internal resistance of 50\ \Omega and gives a full-scale deflection … — JAMB UTME

A galvanometer has an internal resistance of $50\ \Omega$ and gives a full-scale deflection for a current of $2.0\text{ mA}$ . What multiplier resistance must be connected in series with the galvanometer to convert it into a voltmeter capable of measuring potential differences up to $10.0\text{ V}$ ?

$4950\ \Omega$ Answer
B
$5000\ \Omega$
C
$5050\ \Omega$
D
$450\ \Omega$

Answer

The required multiplier resistance is

4950\ \Omega

To convert a galvanometer to a voltmeter, a high-resistance multiplier

R_m

is connected in series. The maximum voltage

V

measured by the voltmeter is given by

V = I_g(G + R_m)

. Substituting

V = 10.0\text{ V}

I_g = 0.002\text{ A}

, and

G = 50\ \Omega

, we find

R_m = \frac{10.0}{0.002} - 50 = 4950\ \Omega

Step-by-Step Solution

Convert given values to standard SI units.

Galvanometer resistance

G = 50\ \Omega

, full-scale current

I_g = 2.0\text{ mA} = 0.002\text{ A}

, maximum voltage

V = 10.0\text{ V}

Electric current must be expressed in amperes (A) for standard circuit calculations.

Apply the series multiplier formula for a voltmeter.

V = I_g(G + R_m) \implies 10.0 = 0.002 \times (50 + R_m)

Converting a galvanometer to a voltmeter requires connecting a high resistance

R_m

in series so that the total voltage drop equals

V

Solve for the multiplier resistance

R_m

50 + R_m = \frac{10.0}{0.002} = 5000 \implies R_m = 5000 - 50 = 4950\ \Omega

Subtracting the internal resistance

G

from total resistance gives the value of the multiplier resistor alone.

Key Concept

Galvanometer Conversion to Voltmeter

Estimated Time:1m 30s

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