An alternating voltage E = 200 sqrt{2} sin(10 | vedclass

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

Question

(B) The given alternating voltage is $E = E_0 \sin(\omega t)$, where $E_0 = 200 \sqrt{2} 	ext{ V}$ and $\omega = 100 	ext{ rad/s}$.The capacitive re

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) The given alternating voltage is $E = E_0 \sin(\omega t)$, where $E_0 = 200 \sqrt{2} 	ext{ V}$ and $\omega = 100 	ext{ rad/s}$.The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{100 	imes 1 	imes 10^{-6}} = 10^4 \Omega$.The peak current is $I_0 = \frac{E_0}{X_C} = \frac{200 \sqrt{2}}{10^4} = 2 \sqrt{2} 	imes 10^{-2} 	ext{ A}$.The $A.C.$ ammeter measures the root mean square $(RMS)$ current, $I_{	ext{rms}} = \frac{I_0}{\sqrt{2}}$.$I_{	ext{rms}} = \frac{2 \sqrt{2} 	imes 10^{-2}}{\sqrt{2}} = 2 	imes 10^{-2} 	ext{ A} = 20 	ext{ mA}$.

An alternating voltage $E = 200 \sqrt{2} \sin(100 t) \text{ V}$ is connected to a $1 \mu\text{F}$ capacitor through an $A.C.$ ammeter. The reading of the ammeter shall be: (in $\text{ mA}$)

Similar Questions

For high frequency,a capacitor offers

$A$ $15.0 \; \mu F$ capacitor is connected to a $220 \; V, 50 \; Hz$ source. Find the capacitive reactance and the current ($rms$ and peak) in the circuit. If the frequency is doubled,what happens to the capacitive reactance and the current?

An $A.C.$ source is connected across a pure inductor. Which one of the following figures shows the correct phase relation between the current and $e.m.f.$?

In a purely resistive $ac$ circuit,the current

For obtaining wattless current . . . . . . is connected with ac supply.

Vedclass Test Series

Exam Paper Generator

Online Exam Module