A battery with an electromotive force (e.m.f.) of 12.0\text{ V} and an internal resistance… — JAMB UTME

A battery with an electromotive force (e.m.f.) of $12.0\text{ V}$ and an internal resistance of $1.0\text{ }\Omega$ is connected across a series combination of a $2.0\text{ }\Omega$ resistor and a $3.0\text{ }\Omega$ resistor. What is the potential difference across the terminals of the battery?

$10.0\text{ V}$ Answer
B
$12.0\text{ V}$
C
$2.0\text{ V}$
D
$6.0\text{ V}$

Answer

10.0\text{ V}

The total resistance of the circuit is the sum of the external load resistors and the internal resistance (

2.0 + 3.0 + 1.0 = 6.0\text{ }\Omega

). The total current is

I = \frac{12.0\text{ V}}{6.0\text{ }\Omega} = 2.0\text{ A}

. The potential difference across the terminals of the battery is the voltage across the external load

V = I \times R_{\text{ext}} = 2.0\text{ A} \times 5.0\text{ }\Omega = 10.0\text{ V}

(or

E - Ir = 12.0 - 2.0 = 10.0\text{ V}

Step-by-Step Solution

Calculate total external resistance in series

R_{\text{ext}} = 2.0\text{ }\Omega + 3.0\text{ }\Omega = 5.0\text{ }\Omega

Resistors in series add directly to give the total external load resistance.

Calculate total circuit resistance including internal resistance

R_{\text{total}} = R_{\text{ext}} + r = 5.0\text{ }\Omega + 1.0\text{ }\Omega = 6.0\text{ }\Omega

The cell's internal resistance is in series with the external circuit.

Determine the total current drawn from the battery

I = \frac{E}{R_{\text{total}}} = \frac{12.0\text{ V}}{6.0\text{ }\Omega} = 2.0\text{ A}

Applying Ohm's law to the entire circuit.

Calculate terminal potential difference across the battery

V = E - Ir = 12.0\text{ V} - (2.0\text{ A} \times 1.0\text{ }\Omega) = 10.0\text{ V}

Terminal voltage is the e.m.f. minus the lost volts across internal resistance (or equivalently

V = I \times R_{\text{ext}}

Key Concept

Terminal potential difference vs. electromotive force (e.m.f.) and lost volts

Estimated Time:1m 30s

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