The voltage of an ac supply varies with time | vedclass

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

Question

(D) Given the voltage equation: $V = 120 \sin(100\pi t) \cos(100\pi t)$.Using the trigonometric identity $\sin(2	heta) = 2 \sin 	heta \cos 	heta$,w

Vedclass · Accepted Answer

$60 \, V, 100 \, Hz$

Vedclass · Answer

(D) Given the voltage equation: $V = 120 \sin(100\pi t) \cos(100\pi t)$.Using the trigonometric identity $\sin(2	heta) = 2 \sin 	heta \cos 	heta$,we can rewrite the equation as:$V = 60 	imes (2 \sin(100\pi t) \cos(100\pi t))$$V = 60 \sin(200\pi t)$.Comparing this with the standard form $V = V_{\max} \sin(\omega t)$,we get:Maximum voltage $V_{\max} = 60 \, V$.Angular frequency $\omega = 200\pi \, rad/s$.Since $\omega = 2\pi 
u$,the frequency $
u$ is:$
u = \frac{\omega}{2\pi} = \frac{200\pi}{2\pi} = 100 \, Hz$.Thus,the maximum voltage is $60 \, V$ and the frequency is $100 \, Hz$.

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

The voltage of an $ac$ supply varies with time $(t)$ as $V = 120\sin(100\pi t)\cos(100\pi t)$. The maximum voltage and frequency respectively are:

Similar Questions

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

The ratio of the peak value to the root mean square (r.m.s.) value of an alternating current is:

$AC$ measuring instruments measure

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

The average value of the potential difference $V$ shown in the figure is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module