PhysicsEasy94×since 2002Q5926An alternating voltage V(t)=220sin100πtV(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:A7.2 msB3.3 msC5 msD2.2 msCheck answerSkip