A galvanometer with an internal resistance of 5\text{ }\Omega produces a full-scale deflec… — JAMB UTME

A galvanometer with an internal resistance of $5\text{ }\Omega$ produces a full-scale deflection when a current of $10\text{ mA}$ flows through it. Calculate the shunt resistance, in ohms, required to convert this galvanometer into an ammeter capable of measuring currents up to $50\text{ mA}$ .

Answer: 1.25 Ω

Answer

The required shunt resistance is

1.25\text{ }\Omega

To extend the range of a galvanometer, a shunt resistor

S

is placed in parallel with it. The potential difference across the galvanometer equals the potential difference across the shunt:

I_g R_g = (I - I_g) S

. Substituting the given values

R_g = 5\text{ }\Omega

I_g = 10\text{ mA}

, and maximum current

I = 50\text{ mA}

yields

10\text{ mA} \times 5\text{ }\Omega = (50\text{ mA} - 10\text{ mA}) \times S

, solving to

S = \frac{50}{40} = 1.25\text{ }\Omega

Step-by-Step Solution

Find the current passing through the parallel shunt resistor

I_s = 40\text{ mA}

By Kirchhoff's current law, the total maximum current divides into the galvanometer current and the shunt current (

I = I_g + I_s

Calculate the required shunt resistance

S

S = 1.25\text{ }\Omega

Because the galvanometer and shunt resistor are connected in parallel, the potential difference across both branches is equal (

I_g R_g = I_s S

Key Concept

Conversion of a galvanometer to an ammeter using a low-resistance shunt in parallel

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