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:

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(2) 3.3 ms

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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:

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An alternating voltage V(t)=220sin⁡100πtV(t) = 220 \sin 100 \pi tV(t)=220sin100πt volt is applied to a purely resistive load of 50Ω50 \Omega50Ω. The time taken for the current to rise from half of the peak value to the peak value is:

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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: