An electric lamp of resistance 10\,\Omega is connected to a cell of electromotive force (e… — JAMB UTME

An electric lamp of resistance $10\,\Omega$ is connected to a cell of electromotive force (e.m.f.) $12\,\text{V}$ and internal resistance $2\,\Omega$ . What is the electrical power dissipated as heat inside the cell?

$2.0\,\text{W}$ Answer
B
$2.88\,\text{W}$
C
$10.0\,\text{W}$
D
$12.0\,\text{W}$

Answer

The electrical power dissipated inside the cell is

2.0\,\text{W}

The total opposition to current in the circuit includes both the external lamp resistance and the internal resistance of the cell (

10\,\Omega + 2\,\Omega = 12\,\Omega

). This yields a circuit current of

1.0\,\text{A}

. Using Joule's law of heating (

P = I^2 r

), the power lost specifically inside the cell is

(1.0)^2 \times 2 = 2.0\,\text{W}

Step-by-Step Solution

Calculate the total resistance of the circuit.

R_{\text{total}} = R + r = 10\,\Omega + 2\,\Omega = 12\,\Omega

The internal resistance of the cell is in series with the external load resistance.

Determine the current flowing through the circuit.

I = \frac{E}{R_{\text{total}}} = \frac{12\,\text{V}}{12\,\Omega} = 1.0\,\text{A}

By Ohm's law applied to a complete circuit, current equals total e.m.f. divided by total resistance.

Calculate the power dissipated as heat in the internal resistance.

P_{\text{internal}} = I^2 r = (1.0\,\text{A})^2 \times 2\,\Omega = 2.0\,\text{W}

Joule heating power in a resistor is given by

P = I^2 r

Key Concept

Electrical Power Dissipation and Internal Resistance

Estimated Time:1m 30s

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