The voltage of an ac source varies with time | vedclass

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

Question

(B) Given the equation $V = 100 \sin(100 \pi t) \cos(100 \pi t)$.Using the trigonometric identity $\sin(2	heta) = 2 \sin 	heta \cos 	heta$,we can r

Vedclass · Accepted Answer

The peak voltage of the source is $50 \, V$.

Vedclass · Answer

(B) Given the equation $V = 100 \sin(100 \pi t) \cos(100 \pi t)$.Using the trigonometric identity $\sin(2	heta) = 2 \sin 	heta \cos 	heta$,we can rewrite the equation as:$V = 50 	imes [2 \sin(100 \pi t) \cos(100 \pi t)]$$V = 50 \sin(200 \pi t)$.Comparing this with the standard form $V = V_0 \sin(\omega t)$,we get:Peak voltage $V_0 = 50 \, V$.Angular frequency $\omega = 200 \pi \, rad/s$.Since $\omega = 2 \pi f$,we have $200 \pi = 2 \pi f$,which gives $f = 100 \, Hz$.Therefore,the peak voltage is $50 \, V$.

The voltage of an $ac$ source varies with time according to the equation $V = 100 \sin(100 \pi t) \cos(100 \pi t)$,where $t$ is in $seconds$ and $V$ is in $volts$. Then:

Similar Questions

If $f: R \rightarrow C$ is defined by $f(x)=e^{2 i x}$ for $x \in R$,then $f$ is (where $C$ denotes the set of all complex numbers)

Which of the following is not a group of cryptogams (non-flowering plants)?

If $f(x) = \frac{x}{1+x}$ and $g(x) = f(f(x))$,then $g^{\prime}(x)$ is equal to

Which of the following options is correct for an object having a straight-line motion represented by the following graph?

The angle between the asymptotes of the hyperbola $x^2-3y^2=3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module