Question 25: AC fundamentals

Question

A bulb of resistance 280 Ohm is supplied with a voltage $\mathrm{V}=400 \sin \pi \mathrm{t}$. The peak current is

Accepted Answer

The supply voltage is of the form
$$V(t)=V_{\!m}\sin(\omega t)$$
with $V_{\!m}=400\rm\,V$.  Since the bulb is purely resistive ($R=280\Omega$), the peak current is
$$ I_{\!m}=\frac{V_{\!m}}{R} =\frac{400}{280}\approx1.43\rm\,A. $$
Answer: 1.43 A (Option D).

Answer

2.22 A

Answer

2.02 A

Answer

1.11 A

Answer

1.43 A

AC fundamentals

A bulb of resistance 280 Ohm is supplied with a voltage \mathrm{V}=400 \sin \pi \mathrm{t}. The peak current is

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