An alternating voltage V(t) = 220 sin(100 pi | vedclass

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(B) The given voltage is $V(t) = 220 \sin(100 \pi t)$. Since the load is purely resistive,the current $I(t)$ is in phase with the voltage: $I(t) = I_0

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$3.3$

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(B) The given voltage is $V(t) = 220 \sin(100 \pi t)$. Since the load is purely resistive,the current $I(t)$ is in phase with the voltage: $I(t) = I_0 \sin(100 \pi t)$,where $I_0 = V_0 / R = 220 / 50 = 4.4 \ A$.We need the time taken for the current to rise from $I_0 / 2$ to $I_0$.At $t_1$,$I(t_1) = I_0 \sin(100 \pi t_1) = I_0 / 2 \implies 100 \pi t_1 = \pi / 6 \implies t_1 = 1 / 600 \ s$.At $t_2$,$I(t_2) = I_0 \sin(100 \pi t_2) = I_0 \implies 100 \pi t_2 = \pi / 2 \implies t_2 = 1 / 200 \ s$.The time taken is $\Delta t = t_2 - t_1 = 1/200 - 1/600 = (3 - 1) / 600 = 2 / 600 = 1 / 300 \ s$.$\Delta t = 0.00333 \ s = 3.33 \ ms$.

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: (in $ms$)

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