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 $100 \, Hz$.

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

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

The frequency of $AC$ mains in India is ....... $Hz$.

An alternating voltage is represented by $V = 80 \sin(100 \pi t) \cos(100 \pi t) \text{ V}$. The peak voltage is (in $\text{ V}$)

$A$ light bulb is rated at $100\;W$ for a $220\;V$ supply. Find
$(a)$ the resistance of the bulb;
$(b)$ the peak voltage of the source; and
$(c)$ the rms current through the bulb.

$100\text{ W}$ रेटेड एक प्रकाश बल्ब $220\text{ V}$, $50\text{ Hz}$ के $AC$ स्रोत से जुड़ा है। बल्ब से प्रवाहित होने वाली $rms$ धारा है: (in $text{ A}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module