For a domestic AC supply of 220 V at 50 tex | vedclass

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

Question

(A) The root mean square voltage is given as $V_{rms} = 220 \ V$. The peak voltage $V_0$ is related to $V_{rms}$ by the formula $V_0 = V_{rms} \sqrt{2

Vedclass · Accepted Answer

$V(t) = 220 \sqrt{2} \cos(100 \pi t)$

Vedclass · Answer

(A) The root mean square voltage is given as $V_{rms} = 220 \ V$. The peak voltage $V_0$ is related to $V_{rms}$ by the formula $V_0 = V_{rms} \sqrt{2}$. Therefore, $V_0 = 220 \sqrt{2} \ V$. The angular frequency $\omega$ is given by $\omega = 2 \pi f$, where $f = 50 \ 	ext{Hz}$. Thus, $\omega = 2 \pi 	imes 50 = 100 \pi \ 	ext{rad/s}$. The instantaneous potential difference is given by $V(t) = V_0 \cos(\omega t)$. Substituting the values, we get $V(t) = 220 \sqrt{2} \cos(100 \pi t)$.

For a domestic $AC$ supply of $220 \ V$ at $50 \ \text{cycles per sec}$, the potential difference between the terminals of a two-pin electric outlet in a room is given by

Similar Questions

What are $DC$ signals and $AC$ signals? Why do we prefer an $AC$ signal?

$A$ sinusoidal voltage produced by an $AC$ generator at any instant $t$ is given by the equation $V = 311 \sin(314t)$. The $rms$ value of voltage and frequency are respectively:

Two sinusoidal voltages of the same frequency are shown in the diagram. What is the frequency, and the phase relationship between the voltages? (Frequency in $Hz$, Phase lead of $N$ over $M$ in radians)

$A$ resistance of $20 \Omega$ is connected to a source of an alternating potential $V = 200 \sin (10 \pi t)$. If $t$ is the time taken by the current to change from the peak value to rms value,then $t$ is (in seconds).

$A$ bulb of resistance $280 \Omega$ is supplied with a $200 V$ $AC$ supply. What is the peak current?

Vedclass Test Series

Exam Paper Generator

Online Exam Module