An alternating voltage v(t) = 220 sin 100 π t volt is applied to a purely resistance 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 2019 Physics question

JEE Main 2019 · 8 Apr, Morning shift · Q5878

JEE Main 20198 Apr, Morning shiftPhysicsHardElectromagnetic Induction and Alternating Currents149× since 2002

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 :

JEE Main 2019 · 8 Apr, Morning shift · Q5878

JEE Main 20198 Apr, Morning shiftPhysicsHardElectromagnetic Induction and Alternating Currents149× since 2002

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 :