An alternating voltage v(t) = 220 sin 100 nt 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 :

Question

An alternating voltage v(t) = 220 sin 100 nt 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 : (A) 2.2 ms (B) 5 ms (C) 3.3 ms (D) 7.2 ms

Tardigrade Admin · Accepted Answer

V(t) = 220 sin (100 π t) volt time taken, t = (θ/ω) = ((π/3)/100π) = (1/300) sec = 3.3 ms

Q. An alternating voltage $v (t) = 220$ $s in 100 n t$ 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 :

2.2 ms

5 ms

3.3 ms

7.2 ms

$V (t) = 220 sin (100 π t) v o lt$
time taken,
$t = \frac{θ}{ω} = \frac{\frac{π}{3}}{100 π} = \frac{1}{300} sec$
$= 3.3 m s$

Q. An alternating voltage v(t)=220 sin100nt volt is applied to a purely resistance load of 50Ω. The time taken for the current to rise from half of the peak value to the peak value is :