For an AC current I = 50 cos(100t + 45^{circ} | vedclass

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

Question

(D) The given equation for the alternating current is $I = I_m \cos(\omega t + \phi)$,where $I_m$ is the peak current.Comparing this with the given eq

Vedclass · Accepted Answer

$25 \sqrt{2}$

Vedclass · Answer

(D) The given equation for the alternating current is $I = I_m \cos(\omega t + \phi)$,where $I_m$ is the peak current.Comparing this with the given equation $I = 50 \cos(100t + 45^{\circ}) \ A$,we find the peak current $I_m = 50 \ A$.The root mean square current $I_{rms}$ is related to the peak current $I_m$ by the formula $I_{rms} = \frac{I_m}{\sqrt{2}}$.Substituting the value of $I_m$,we get $I_{rms} = \frac{50}{\sqrt{2}}$.To simplify,multiply the numerator and denominator by $\sqrt{2}$: $I_{rms} = \frac{50 \sqrt{2}}{2} = 25 \sqrt{2} \ A$.Therefore,the correct option is $D$.

For an $AC$ current $I = 50 \cos(100t + 45^{\circ}) \ A$. The value of $I_{rms} =$ . . . . . . $A$.

Similar Questions

What will be the $r.m.s.$ value of the given $A.C.$ voltage over one complete cycle?

$A$ resistance of $20 \Omega$ is connected to an alternating current source of $110 V$. If the frequency of the $A.C.$ source is $50 Hz$,then the time taken by the current to change from its maximum value to the $R.M.S.$ value is

The mean and $rms$ value of an alternating voltage for a half cycle,as shown in the figure,are respectively:

Find the time required for $50\,Hz$ alternating current to change its value from zero to maximum value.

An $AC$ source rated $220\, V, 50\, Hz$ is connected to a resistor. The time taken by the current to change from its maximum value to the $rms$ value is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module