Question 220: AC fundamentals

Question

An alternating voltage v(t) = 220 sin 100 $$\pi $$t volt
is applied to a purely resistance 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

In an AC resistive circuit, current and voltage are in phase.
So, $$I = {V \over R} \Rightarrow I = {{220} \over {50}}\sin (100\pi t)$$ ..... (i)
$$	herefore$$ Time period of one complete cycle of current is
$$T = {{2\pi } \over \omega } = {{2\pi } \over {100\pi }} = {1 \over {50}}s$$
So, current reaches its maximum value at
$${t_1} = {T \over 4} = {1 \over {200}}s$$
When current is half of its maximum value, then from Eq.(i), we have
$$I = {{{I_{\max }}} \over 2} = {I_{\max }}\sin (100\pi {t_2})$$
$$ \Rightarrow \sin (100\pi {t_2}) = {1 \over 2} \Rightarrow 100\pi {t_2} = {{5\pi } \over 6}$$
So, instantaneous time at which current is half of maximum value is $${t_2} = {1 \over {120}}s$$
Hence, time duration in which current reaches half of its maximum value after reaching maximum value is
$$\Delta t = {t_2} - {t_1} = {1 \over {120}} - {1 \over {200}} = {1 \over {300}}s = 3.3$$ ms

Answer

5 ms

Answer

2.2 ms

Answer

3.3 ms

Answer

7.2 ms

AC fundamentals

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

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