What is the sum of the instantaneous current | vedclass

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

Question

(A) The instantaneous current in an $AC$ circuit is given by $I(t) = I_{max} \sin(\omega t)$.To find the sum (or integral) of the instantaneous curren

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) The instantaneous current in an $AC$ circuit is given by $I(t) = I_{max} \sin(\omega t)$.To find the sum (or integral) of the instantaneous current values over one complete cycle of period $T$,we calculate the integral $\int_{0}^{T} I(t) dt$.$\int_{0}^{T} I_{max} \sin(\omega t) dt = I_{max} [-\frac{\cos(\omega t)}{\omega}]_{0}^{T}$.Since $\omega = \frac{2\pi}{T}$,we have $\omega T = 2\pi$.Substituting the limits: $I_{max} [-\frac{\cos(2\pi)}{\omega} - (-\frac{\cos(0)}{\omega})] = I_{max} [-\frac{1}{\omega} + \frac{1}{\omega}] = 0$.Therefore,the sum of the instantaneous current values over one complete $AC$ cycle is $0$.

$(A) \ x_0 \sin \omega t$	$(i) \ x_0$
$(B) \ x_0 \sin \omega t \cos \omega t$	$(ii) \ \frac{x_0}{\sqrt{2}}$
$(C) \ x_0 \sin \omega t + x_0 \cos \omega t$	$(iii) \ \frac{x_0}{2 \sqrt{2}}$

What is the sum of the instantaneous current values over one complete $AC$ cycle?

Similar Questions

What is the average value of the $AC$ voltage over a complete cycle?

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

In an $AC$ generator, induced emf $\varepsilon = 0$ at $t = 0$, then its value . . . . . .

The a.c. source of e.m.f. with instantaneous value $e$ is given by $e = 200 \sin(50t) \text{ V}$. The r.m.s. value of current in a circuit of resistance $50 \ \Omega$ is: (in $\text{ A}$)

Match the following current $\text{r.m.s.}$ values:
$(A) \ x_0 \sin \omega t$ $(i) \ x_0$
$(B) \ x_0 \sin \omega t \cos \omega t$ $(ii) \ \frac{x_0}{\sqrt{2}}$
$(C) \ x_0 \sin \omega t + x_0 \cos \omega t$ $(iii) \ \frac{x_0}{2 \sqrt{2}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module