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WAEC Physics 2010 Paper

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Question : 36

Total: 50

Use the diagram above to calculate the current, I.

0.60A
0.97A
1.03A
5.00A

Solution:

Concept:

The circuit has a

6\Omega

and a

4\Omega

resistor connected in parallel, and this parallel combination is connected in series with a

10\Omega

resistor.

Explanation:

For the parallel combination, use:

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

\frac{1}{R_p} = \frac{2+3}{12} = \frac{5}{12}

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

Total resistance of the circuit is:

R_T = R_p + 10 = 2.4 + 10 = 12.4\Omega

Using Ohm's law:

I = \frac{V}{R_T} = \frac{12}{12.4}

I = 0.9677A \approx 0.97A

Answer:

The current

I

is

0.97A

.

Correct option: B.

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