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WAEC Physics 2010 Paper

Question : 37

Total: 50

Solution:

Concept:

Use the equivalent resistance of the parallel combination and apply Ohm's law.

Explanation:

From the diagram, the parallel group of

6\Omega

and

4\Omega

is connected in series with a

0.7\Omega

resistor.

Equivalent parallel resistance:

\frac{1}{R_p} = \frac{1}{6} + \frac{1}{4} = \frac{5}{12}

R_p = \frac{12}{5} = 2.4\Omega

Total circuit resistance:

R_T = 0.7 + 2.4 = 3.1\Omega

Current drawn from the

3.0\text{ V}

supply:

I = \frac{3.0}{3.1} \approx 0.97\text{ A}

Potential difference across the parallel resistors:

V = IR_p = 0.97 \times 2.4 \approx 2.3\text{ V}

Answer:

2.3\text{ V}

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