Concept:We must derive the sinusoidal AC current equation \( I(t) = I_0 \sin(\omega t) \) using Ohm's Law and angular conversion.
Formula:$$V_0 = V_{\text{rms}} \sqrt{2}$$
$$I_0 = \frac{V_0}{R}$$
$$\omega = 2\pi f$$
Solution:Given values:
- \( V_{\text{rms}} = 220\text{ V} \)
- \( f = 50\text{ Hz} \)
- \( R = 50\text{ k}\Omega = 50,000\ \Omega \)
1. Calculate the peak voltage \( V_0 \):
$$V_0 = 220 \times \sqrt{2} \approx 220 \times 1.414 = 311.1\text{ V}$$
2. Calculate the peak current \( I_0 \):
$$I_0 = \frac{311.1\text{ V}}{50,000\ \Omega} = 0.00622\text{ A} = 6.22\text{ mA}$$
3. Calculate the angular frequency \( \omega \):
$$\omega = 2 \pi \times 50 = 100 \pi \approx 314\text{ rad/s}$$
4. Formulate the equation:
$$I(t) = 6.2 \sin(314t)\text{ mA}$$
Why other options are incorrect:- Equations with 4.4 mA incorrectly use the RMS voltage (\( 220/50 = 4.4 \)) directly instead of computing the peak current \( I_0 \).
- Equations with 157t use an incorrect angular frequency (calculating \( \pi f \) instead of \( 2\pi f \)).
Quality & Fidelity Assurance:
Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.