The r.m.s. value of an AC of 50 Hz is 10 A. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The time period $T$ of the alternating current is given by $T = \frac{1}{
u} = \frac{1}{50} \ s = 0.02 \ s$.The time taken to reach the maximum v

Vedclass · Accepted Answer

$5 	imes 10^{-3} \ s$ and $14.14 \ A$

Vedclass · Answer

(D) The time period $T$ of the alternating current is given by $T = \frac{1}{
u} = \frac{1}{50} \ s = 0.02 \ s$.The time taken to reach the maximum value from zero is one-fourth of the time period:$t = \frac{T}{4} = \frac{0.02}{4} = 0.005 \ s = 5 	imes 10^{-3} \ s$.The peak value of the current $(i_0)$ is related to the $r.m.s.$ value $(i_{rms})$ by the formula:$i_0 = i_{rms} 	imes \sqrt{2} = 10 	imes 1.414 = 14.14 \ A$.Therefore,the time taken is $5 	imes 10^{-3} \ s$ and the peak value is $14.14 \ A$.

The $r.m.s.$ value of an $AC$ of $50 \ Hz$ is $10 \ A$. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be:

Similar Questions

The current in a circuit varies with time as $I = 2 \sqrt{t}$. The $rms$ value of the current for the interval $t = 2 \, s$ to $t = 4 \, s$ is:

An alternating current is given by the equation $i = i_1 \cos \omega t + i_2 \sin \omega t$. The r.m.s. current is given by

If an alternating voltage is represented as $E = 141 \sin(628 t)$,then the rms value of the voltage and the frequency are respectively:

Alternating voltage:

$A$ group of lamps having a total power rating of $1000 \,W$ is supplied by an $AC$ voltage of $E = 200 \sin(310t + 60^{\circ})$. The $r.m.s.$ value of the current flowing through the circuit is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module