A cell of electromotive force E = 6.0\text{ V} and internal resistance r = 1.0\text{ }\Ome… — JAMB UTME

A cell of electromotive force $E = 6.0\text{ V}$ and internal resistance $r = 1.0\text{ }\Omega$ is connected in series with a resistor $R$ and a shunted galvanometer. The galvanometer has a resistance of $90\text{ }\Omega$ and produces a full-scale deflection for a current of $2.0\text{ mA}$ . If the shunt resistance connected across the galvanometer is $10\text{ }\Omega$ , calculate the value of the series resistor $R$ , in ohms $(\Omega)$ , required for the galvanometer to show full-scale deflection.

Answer: 290 Ω

Answer

The required value of the series resistor

R

290\text{ }\Omega

At full-scale deflection, a current of

2.0\text{ mA}

passes through the galvanometer, resulting in a potential difference of

V_g = 2.0 \times 10^{-3} \times 90 = 0.18\text{ V}

. Since the shunt is connected in parallel with the galvanometer, the current through the shunt is

I_s = \frac{0.18}{10} = 0.018\text{ A} = 18\text{ mA}

. Thus, the total current supplied by the cell is

I = 2\text{ mA} + 18\text{ mA} = 20\text{ mA} = 0.02\text{ A}

. The total equivalent resistance of the shunted galvanometer is

R_p = \frac{90 \times 10}{90 + 10} = 9.0\text{ }\Omega

. Applying Ohm's law to the total loop including internal resistance

r

, we have

E = I(R + R_p + r) \implies 6.0 = 0.02(R + 9.0 + 1.0) \implies R + 10.0 = 300 \implies R = 290\text{ }\Omega

Step-by-Step Solution

Calculate voltage across the galvanometer at full-scale deflection

V_g = 0.18\text{ V}

The potential difference across parallel branches is equal, and for full-scale deflection

I_g = 2.0\text{ mA}

Calculate the current passing through the shunt resistor

I_s = 18.0\text{ mA} = 0.018\text{ A}

Using Ohm's law across the shunt resistor

S = 10\text{ }\Omega

with

V_s = V_g = 0.18\text{ V}

Calculate the total circuit current provided by the cell

I = 20.0\text{ mA} = 0.020\text{ A}

By Kirchhoff's current law, the main current splits between the galvanometer and shunt.

Find the equivalent resistance of the shunted galvanometer and total circuit resistance

R_p = 9.0\text{ }\Omega

and total circuit resistance

R_{\text{total}} = 300\text{ }\Omega

Parallel resistance formula gives

R_p = 9.0\text{ }\Omega

, and

R_{\text{total}} = \frac{E}{I} = \frac{6.0\text{ V}}{0.020\text{ A}} = 300\text{ }\Omega

Solve for the unknown external series resistance

R

R = 290\text{ }\Omega

R_{\text{total}} = R + r + R_p \implies 300 = R + 1.0 + 9.0 \implies R = 290\text{ }\Omega

Key Concept

Galvanometer Shunting and Electric Circuit Analysis with Internal Resistance

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