An electric cell of electromotive force E and internal resistance r is connected across a … — JAMB UTME

An electric cell of electromotive force $E$ and internal resistance $r$ is connected across a parallel combination of two resistors with resistances $6.0\text{ }\Omega$ and $12.0\text{ }\Omega$ . The potential difference across the parallel combination is $4.0\text{ V}$ . When the $12.0\text{ }\Omega$ resistor is removed from the circuit, the current supplied by the cell becomes $0.75\text{ A}$ . What is the internal resistance $r$ of the cell in ohms?

Answer: 2 \Omega

Answer

The internal resistance of the cell is

2.0\text{ }\Omega

Analyzing the circuit under both states yields two simultaneous equations for the e.m.f.

E

in terms of internal resistance

r

E = 4.0 + 1.0r

and

E = 4.5 + 0.75r

. Solving these equations gives

r = 2.0\text{ }\Omega

Step-by-Step Solution

Calculate the equivalent resistance of the parallel resistor network.

R_p = 4.0\text{ }\Omega

Using the parallel resistor formula:

R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{6.0 \times 12.0}{6.0 + 12.0} = 4.0\text{ }\Omega

Determine the initial total current delivered by the cell.

I_1 = 1.0\text{ A}

From the terminal voltage across the parallel load:

I_1 = \frac{V_1}{R_p} = \frac{4.0\text{ V}}{4.0\text{ }\Omega} = 1.0\text{ A}

Formulate the e.m.f. equation for the initial circuit state.

E = 4.0 + 1.0r

Applying the equation

E = V + Ir

gives

E = 4.0 + (1.0)r

Formulate the e.m.f. equation after removing the

12.0\text{ }\Omega

resistor.

E = 4.5 + 0.75r

With external load

R_2 = 6.0\text{ }\Omega

and current

I_2 = 0.75\text{ A}

E = I_2(R_2 + r) = 0.75(6.0 + r) = 4.5 + 0.75r

Solve for the internal resistance

r

by equating the two e.m.f. expressions.

r = 2.0\text{ }\Omega

Equating the two expressions for

E

4.0 + 1.0r = 4.5 + 0.75r \implies 0.25r = 0.50 \implies r = 2.0\text{ }\Omega

Key Concept

Internal Resistance and Multi-State Circuit Analysis

Estimated Time:2m 0s

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