An alternating voltage V(t)=220 sin 100 π t volt is applied to a purely resistive load of 50 Ω. The time taken for the current to rise from half of the peak value to the peak value is:

Electromagnetic Induction and Alternating Currents: JEE Main 2024 Physics question

JEE Main 2024 · 30 Jan, Evening shift · Q5926

JEE Main 202430 Jan, Evening shiftPhysicsEasyElectromagnetic Induction and Alternating Currents149× since 2002

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:

JEE Main 2024 · 30 Jan, Evening shift · Q5926

JEE Main 202430 Jan, Evening shiftPhysicsEasyElectromagnetic Induction and Alternating Currents149× since 2002

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: