A resistance of 20 , Omega is connected to a | vedclass

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

Question

(D) The peak value of current is $i_0$ and the r.m.s. value is $i_{rms} = \frac{i_0}{\sqrt{2}}$.Given the alternating current equation $i = i_0 \sin(1

Vedclass · Accepted Answer

$2.5 	imes 10^{-3} \, 	ext{s}$

Vedclass · Answer

(D) The peak value of current is $i_0$ and the r.m.s. value is $i_{rms} = \frac{i_0}{\sqrt{2}}$.Given the alternating current equation $i = i_0 \sin(100 \pi t)$.At peak value,the phase $\omega t_1 = \frac{\pi}{2}$,so $t_1 = \frac{\pi}{2 	imes 100 \pi} = \frac{1}{200} \, 	ext{s}$.At r.m.s. value,$i = \frac{i_0}{\sqrt{2}}$,so $\sin(100 \pi t_2) = \frac{1}{\sqrt{2}} = \sin(\frac{\pi}{4})$.Thus,$100 \pi t_2 = \frac{\pi}{4} \Rightarrow t_2 = \frac{1}{400} \, 	ext{s}$.The time taken is $\Delta t = t_1 - t_2 = \frac{1}{200} - \frac{1}{400} = \frac{2-1}{400} = \frac{1}{400} \, 	ext{s}$.$\Delta t = 0.0025 \, 	ext{s} = 2.5 	imes 10^{-3} \, 	ext{s}$.

$A$ resistance of $20 \, \Omega$ is connected to a source of an alternating potential $V = 220 \sin(100 \pi t)$. The time taken by the current to change from its peak value to its r.m.s. value is:

Similar Questions

An electric lamp is connected to a $220 V, 50 Hz$ supply. The peak value of the voltage is......$V$.

Alternating voltage:

Find the time required for a $50 \text{ Hz}$ alternating current to reach its $rms$ value from zero. (in $\text{ ms}$)

An alternating voltage is given by $E = 100 \sin \left(\omega t + \frac{\pi}{6}\right) \text{ V}$. The voltage will be maximum for the first time when $t$ is $(T = \text{periodic time})$:

An $AC$ current is given by $I = I_0 + I_1 \sin \omega t$. Then its $rms$ value will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module