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WAEC Physics 2024 Paper

Question : 23

Total: 50

_{\text{rms}}

Solution:

Concept:

A capacitor connected to an AC source offers capacitive reactance, and the rms current is given by

I_{\text{rms}} = \dfrac{V_{\text{rms}}}{X_c}

Explanation:

The capacitive reactance is

X_c = \dfrac{1}{2\pi f C}

Here,

C = 2\ \mu\mathrm{F} = 2 \times 10^{-6}\ \mathrm{F}

and

f = 60\ \mathrm{Hz}

Substitute

\pi = 3.142

X_c = \dfrac{1}{2 \times 3.142 \times 60 \times 2 \times 10^{-6}}

X_c = 1326.12\ \Omega

Now use the AC version of Ohm's law:

I_{\text{rms}} = \dfrac{V_{\text{rms}}}{X_c} = \dfrac{150}{1326.12}

I_{\text{rms}} = 0.113\ \mathrm{A}

Answer:

The rms value of the current is

0.113\ \mathrm{A}

, which is option A.

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