A cell of electromotive force E and internal resistance r supplies a current of 1.2\text{ … — JAMB UTME

A cell of electromotive force $E$ and internal resistance $r$ supplies a current of $1.2\text{ A}$ when connected across a $4.0\ \Omega$ resistor. When this resistor is replaced with a $10.0\ \Omega$ resistor, the current decreases to $0.6\text{ A}$ . Calculate the internal resistance $r$ of the cell in ohms ( $\Omega$ ).

Answer: 2 Ω

Answer

The internal resistance of the cell is

2.0\ \Omega

For a complete circuit, the electromotive force

E

is related to load resistance

R

, internal resistance

r

, and current

I

E = I(R + r)

. Setting up equations for both load conditions gives

E = 1.2(4.0 + r)

and

E = 0.6(10.0 + r)

. Equating these expressions yields

4.8 + 1.2r = 6.0 + 0.6r

, which simplifies to

0.6r = 1.2

, giving an internal resistance of

r = 2.0\ \Omega

Step-by-Step Solution

Formulate the circuit equation for the first load resistance

E = 1.2 \times (4.0 + r) = 4.8 + 1.2r

Using Ohm's law for a complete circuit, the electromotive force equals current times total circuit resistance:

E = I(R_1 + r)

Formulate the circuit equation for the second load resistance

E = 0.6 \times (10.0 + r) = 6.0 + 0.6r

The cell maintains the same internal electromotive force

E

and internal resistance

r

with the new load resistor

R_2

Equate the expressions for

E

and solve for

r

4.8 + 1.2r = 6.0 + 0.6r \implies 0.6r = 1.2 \implies r = 2.0\ \Omega

Since

E

is constant for the cell, setting the two right-hand sides equal yields a single linear equation for the unknown internal resistance

r

Key Concept

Electromotive Force and Internal Resistance

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