A battery with an electromotive force (e.m.f.) of 9.0\text{ V} and an internal resistance … — JAMB UTME

A battery with an electromotive force (e.m.f.) of $9.0\text{ V}$ and an internal resistance of $0.5\text{ }\Omega$ is connected to an external circuit containing two resistors of $4.0\text{ }\Omega$ and $1.5\text{ }\Omega$ connected in series. What is the terminal potential difference across the battery?

$8.25\text{ V}$ Answer
B
$9.00\text{ V}$
C
$0.75\text{ V}$
D
$6.00\text{ V}$

Answer

The terminal potential difference across the battery is

8.25\text{ V}

The correct answer is obtained by finding the total circuit resistance (

6.0\text{ }\Omega

), determining the current (

1.5\text{ A}

), and subtracting the internal voltage drop (

0.75\text{ V}

) from the electromotive force (

9.0\text{ V}

) to give

8.25\text{ V}

Step-by-Step Solution

Calculate the total resistance of the circuit.

R_{\text{total}} = R_1 + R_2 + r = 4.0\text{ }\Omega + 1.5\text{ }\Omega + 0.5\text{ }\Omega = 6.0\text{ }\Omega

The external resistors and the battery's internal resistance are connected in series.

Calculate the total current flowing through the circuit using Ohm's Law.

I = \frac{E}{R_{\text{total}}} = \frac{9.0\text{ V}}{6.0\text{ }\Omega} = 1.5\text{ A}

The total current depends on the electromotive force and total circuit resistance.

Calculate the terminal potential difference across the battery.

V = E - I r = 9.0\text{ V} - (1.5\text{ A} \times 0.5\text{ }\Omega) = 9.0\text{ V} - 0.75\text{ V} = 8.25\text{ V}

Terminal voltage is equal to the e.m.f. minus the potential drop across the internal resistance.

Key Concept

Terminal potential difference and internal resistance in DC electric circuits

Estimated Time:1m 30s

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