A cell supplies a current of 0.6\text{ A} when connected across a 2.0\text{ }\Omega resist… — JAMB UTME

A cell supplies a current of $0.6\text{ A}$ when connected across a $2.0\text{ }\Omega$ resistor. When the resistor is replaced with a $5.0\text{ }\Omega$ resistor, the current supplied by the cell drops to $0.3\text{ A}$ . What is the internal resistance of the cell?

$1.0\text{ }\Omega$ Answer
B
$0.0\text{ }\Omega$
C
$3.0\text{ }\Omega$
D
$0.5\text{ }\Omega$

Answer

The internal resistance of the cell is

1.0\text{ }\Omega

The electromotive force (e.m.f.)

E

of a cell is given by

E = I(R + r)

, where

I

is the current,

R

is external resistance, and

r

is internal resistance. Setting up equations for both cases:

E = 0.6(2.0 + r)

and

E = 0.3(5.0 + r)

. Equating them gives

1.2 + 0.6r = 1.5 + 0.3r

, which simplifies to

0.3r = 0.3

, yielding

r = 1.0\text{ }\Omega

Step-by-Step Solution

Formulate the electromotive force equation for the first circuit setup

E = I_1(R_1 + r) = 0.6(2.0 + r) = 1.2 + 0.6r

The total voltage supplied by the cell equals the total current multiplied by the sum of external and internal resistance.

Formulate the electromotive force equation for the second circuit setup

E = I_2(R_2 + r) = 0.3(5.0 + r) = 1.5 + 0.3r

The cell's electromotive force

E

and internal resistance

r

remain unchanged despite changing the external resistor.

Equate both expressions for electromotive force and solve for internal resistance

1.2 + 0.6r = 1.5 + 0.3r \implies 0.3r = 0.3 \implies r = 1.0\text{ }\Omega

Equating the two expressions allows direct solution for the single unknown variable

r

Key Concept

Electromotive Force and Internal Resistance of Cells

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