An alternating voltage v(t)=220sin100πt volt is applied to a purely resistive load of 50Ω .The time taken for the current to rise from half of the peak value to the peak value is (in milli-second)

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 3.33.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

As V(t)= 220sin100πtSo, I(t)= 22050sin 100πti.e; I= Imsin(100πt)For I= Im t1=π2×1100π=1200secAnd for I= Im2⇒Im2=Imsin(100πt2)⇒t2=1600s∴treq=1200−1600=2600=1300s=3.33ms

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET