Question 147: AC fundamentals

Question

An alternating voltage $$V(t)=220 \sin 100 \pi t$$ 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:

Accepted Answer

Rising half to peak
$$\begin{aligned}
& \mathrm{t}=\mathrm{T} / 6 \
& \mathrm{t}=\frac{2 \pi}{6 \omega}=\frac{\pi}{3 \omega}=\frac{\pi}{300 \pi}=\frac{1}{300}=3.33 \mathrm{~ms}
\end{aligned}$$

Answer

7.2 ms

Answer

3.3 ms

Answer

5 ms

Answer

2.2 ms

AC fundamentals

An alternating voltage $V(t)=220 \sin 100 \pi t volt is applied to a purely resistive load of 50 \Omega$. The time taken for the current to rise from half of the peak value to the peak value is:

Choose the correct answer: