At time t=0 text{ s}, the voltage of an AC ge | vedclass

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

Question

(B) The instantaneous voltage of an $AC$ generator is given by $V = V_m \sin(\omega t)$, where $V_m$ is the peak voltage and $\omega = 2 \pi 
u$ is t

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The instantaneous voltage of an $AC$ generator is given by $V = V_m \sin(\omega t)$, where $V_m$ is the peak voltage and $\omega = 2 \pi 
u$ is the angular frequency.Given: $V_m = 100 	ext{ V}$, $V = 2 	ext{ V}$ at $t = \frac{1}{100 \pi} 	ext{ s}$.Substituting these values into the equation:$2 = 100 \sin\left(2 \pi 
u 	imes \frac{1}{100 \pi}ight)$$2 = 100 \sin\left(\frac{
u}{50}ight)$$\frac{2}{100} = \sin\left(\frac{
u}{50}ight)$Since the angle $\frac{
u}{50}$ is very small, we can use the approximation $\sin(	heta) \approx 	heta$:$\frac{1}{50} = \frac{
u}{50}$$
u = 1 	ext{ Hz}$.

At time $t=0 \text{ s}$, the voltage of an $AC$ generator starts from $0 \text{ V}$ and becomes $2 \text{ V}$ at time $t=\frac{1}{100 \pi} \text{ s}$. The voltage increases up to $100 \text{ V}$, after which it starts to decrease. Find the frequency of the generator. (in $\text{ Hz}$)

Similar Questions

Find the effective value of current $i = 2\sin(100\pi t) + 2\cos(100\pi t + 30^{\circ})$.

If an $AC$ main supply is given to be $220\,V$,the average $emf$ during a positive half cycle will be.....$V$.

$A$ standard filament lamp consumes $100\,W$ when connected to $200\,V$ $AC$ mains supply. The peak current through the bulb will be $........\,A$.

If an alternating current is given by $i = a \sin(\omega t) + b \cos(\omega t)$,then the $rms$ value of the current is

Why is $220 \ V$ $AC$ more dangerous than $220 \ V$ $DC$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module