An alternating voltage \( V=220\sin(100\pi t) \) volt is applied to a purely resistive load of \( 50\,\Omega \). The time taken for the current to rise from half of its peak value to the peak value is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 3.3\,\mathrm{ms} \)
In a purely resistive circuit the current is in phase with the voltage, so \( i=i_m\sin(100\pi t) \). So the current goes from half its peak to its peak in the same time as \( \sin(100\pi t) \) goes from 1/2 to 1. The value of the resistance does not affect this time.
The current is half its peak value when:
\( \sin(100\pi t_1)=1/2 \)
\( \Rightarrow\; 100\pi t_1=\pi/6 \)
\( \displaystyle t_1=\frac{1}{600}\,\mathrm{s} \)
It reaches the peak value when:
\( \sin(100\pi t_2)=1 \)
\( \Rightarrow\; 100\pi t_2=\pi/2 \)
\( \displaystyle t_2=\frac{1}{200}\,\mathrm{s} \)
The time taken is:
\( \begin{aligned} \Delta t &= t_2-t_1 \\ &= \left(\frac{1}{200}-\frac{1}{600}\right)\mathrm{s} \\ &= \frac{1}{300}\,\mathrm{s} \\ &\approx 3.3\,\mathrm{ms} \end{aligned} \)