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WAEC Physics 2020 Paper

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Question : 36

Total: 51

In an electric circuit, an inductor of inductance 0.5 H and resistance

50 \Omega

is connected to an alternating current source of frequency 60 Hz . Calculate the impedance of the circuit.

$50.0 \Omega$
$450.5 \Omega$
$195.0 \Omega$
$1950.1 \Omega$

Solution:

Concept:

In a resistor–inductor

(R\text{-}L)

circuit, the total opposition to current is called impedance, which combines resistance and inductive reactance.

Explanation:

Inductive reactance is given by

X_L = 2\pi f L

.

Substitute

f = 60\text{ Hz}

and

L = 0.5\text{ H}

:

X_L = 2\pi \times 60 \times 0.5 = 60\pi \approx 188.5\ \Omega

.

Impedance is

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

.

With

R = 50\ \Omega

:

Z = \sqrt{50^2 + (60\pi)^2}

.

Z = \sqrt{2500 + 35530.6} = \sqrt{38030.6} \approx 195.0\ \Omega

.

Answer:

The impedance of the circuit is

195.0\ \Omega

, which is Option C.

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