An alternating voltage given as V = 100 sqrt{ | vedclass

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

Question

(A) The given alternating voltage is $V = 100 \sqrt{2} \sin(100t) 	ext{ V}$.Comparing this with the standard form $V = V_m \sin(\omega t)$,we get the

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) The given alternating voltage is $V = 100 \sqrt{2} \sin(100t) 	ext{ V}$.Comparing this with the standard form $V = V_m \sin(\omega t)$,we get the peak voltage $V_m = 100 \sqrt{2} 	ext{ V}$ and angular frequency $\omega = 100 	ext{ rad/s}$.The capacitive reactance is given by $X_C = \frac{1}{\omega C}$.Substituting the values,$X_C = \frac{1}{100 	imes 1 	imes 10^{-6}} = \frac{1}{10^{-4}} = 10^4 	ext{ } \Omega$.The root mean square $(RMS)$ voltage is $V_{	ext{rms}} = \frac{V_m}{\sqrt{2}} = \frac{100 \sqrt{2}}{\sqrt{2}} = 100 	ext{ V}$.The $RMS$ current $I_{	ext{rms}}$ is given by $I_{	ext{rms}} = \frac{V_{	ext{rms}}}{X_C}$.$I_{	ext{rms}} = \frac{100}{10^4} = 10^{-2} 	ext{ A}$.Converting to milliamperes,$I_{	ext{rms}} = 10^{-2} 	imes 10^3 	ext{ mA} = 10 	ext{ mA}$.

An alternating voltage given as $V = 100 \sqrt{2} \sin(100t) \text{ V}$ is applied to a capacitor of $1 \mu\text{F}$. The current reading of the ammeter will be equal to . . . . . . $\text{mA}$.

Similar Questions

Determine the frequency for which a $10$ $\mu$$F$ capacitor has a reactance of $2 \times 10^{-3} \Omega$.

The resistance offered by the inductor $(X_L)$ in an ac circuit is

If we increase the frequency of an $a.c.$ supply,then inductive reactance

$A$ $\frac{50}{\pi} \mu F$ capacitor is connected to a $250 \, V, 50 \, Hz$ $AC$ supply. Then the rms current of the circuit is (in $ \, A$)

The inductive reactance of a coil is $R \ \Omega$. If the inductance of the coil is tripled and the frequency of the $A.C.$ supply is also tripled,then the new inductive reactance will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module