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WAEC Physics 2013 Paper

Question : 36

Total: 46

\Omega

\pi

= 3.142]

Solution:

Concept:

An AC coil offers both resistance and inductive reactance, so the total opposition is the impedance

Z

Explanation:

Given:

L = 0.12\ \mathrm{H}

R = 4\ \Omega

V = 240\ \mathrm{V}

f = 50\ \mathrm{Hz}

\pi = 3.142

First, calculate the inductive reactance:

X_L = 2\pi f L

X_L = 2 \times 3.142 \times 50 \times 0.12

X_L = 37.704\ \Omega

Next, calculate the impedance of the coil:

Z = \sqrt{R^2 + X_L^2}

Z = \sqrt{4^2 + 37.704^2}

Z = \sqrt{16 + 1421.59}

Z = 37.90\ \Omega

Now, use Ohm's law for the AC circuit:

I = \frac{V}{Z}

I = \frac{240}{37.90}

I = 6.33\ \mathrm{A} \approx 6.3\ \mathrm{A}

Answer:

The current through the coil is

6.3\ \mathrm{A}

Correct option: A.

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