Alternating Current Question 10
Question 10 - 2024 (30 Jan Shift 2)
An alternating voltage $\mathrm{V}(\mathrm{t})=220 \sin 100 \pi \mathrm{t}$ volt is applied to a purely resistive load of $50 \Omega$. The time taken for the current to rise from half of the peak value to the peak value is:
(1) $5 \mathrm{~ms}$
(2) $3.3 \mathrm{~ms}$
(3) $7.2 \mathrm{~ms}$
(4) $2.2 \mathrm{~ms}$
Show Answer
Answer: (2)
Solution:
Rising half to peak
$\mathrm{t}=\mathrm{T} / 6$
$\mathrm{t}=\frac{2 \pi}{6 \omega}=\frac{\pi}{3 \omega}=\frac{\pi}{300 \pi}=\frac{1}{300}=3.33 \mathrm{~ms}$
