The reactance of a capacitor of capacitance C | vedclass

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

Question

(A) The capacitive reactance $X_C$ is given by the formula: $X_C = \frac{1}{2\pi 
u C}$.Given: $X_C = 25 \ \Omega$,$
u = \frac{400}{\pi} \ Hz$.Rearr

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(A) The capacitive reactance $X_C$ is given by the formula: $X_C = \frac{1}{2\pi 
u C}$.Given: $X_C = 25 \ \Omega$,$
u = \frac{400}{\pi} \ Hz$.Rearranging the formula for $C$: $C = \frac{1}{2\pi 
u X_C}$.Substituting the values: $C = \frac{1}{2 	imes \pi 	imes (\frac{400}{\pi}) 	imes 25}$.$C = \frac{1}{2 	imes 400 	imes 25} = \frac{1}{20000} \ F$.Converting to $\mu F$: $C = \frac{1}{20000} 	imes 10^6 \ \mu F = \frac{100}{2} \ \mu F = 50 \ \mu F$.

The reactance of a capacitor of capacitance $C \mu F$ for an $ac$ frequency of $\frac{400}{\pi} \ Hz$ is $25 \ \Omega$. The value of $C$ is ..... $\mu F$.

Similar Questions

An alternating voltage $V = 300\sqrt{2} \sin(100t)$ is connected to a $1 \ \mu F$ capacitor through an $AC$ ammeter. The reading of the ammeter will be.....$mA$.

In an $AC$ circuit containing only a capacitor,how does the current relate to the potential difference?

The average power dissipation in a pure capacitance in an $AC$ circuit is

$A$ capacitor of $50 \mu F$ is connected to a power source $V = 220 \sin 50 t$ (where $V$ is in volt and $t$ is in second). Find the value of the $rms$ current (in ampere).

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 $e.m.f.$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module