A parallel plate capacitor having plates of r | vedclass

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

Question

(A) The r.m.s. current $I_{rms}$ in a capacitive circuit is given by $I_{rms} = \frac{V_{rms}}{X_C}$,where $X_C = \frac{1}{\omega C}$ is the capacitiv

Vedclass · Accepted Answer

$6.9 	imes 10^{-6} 	ext{ A}$

Vedclass · Answer

(A) The r.m.s. current $I_{rms}$ in a capacitive circuit is given by $I_{rms} = \frac{V_{rms}}{X_C}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.Substituting $X_C$,we get $I_{rms} = V_{rms} \cdot \omega \cdot C$.Given values are $V_{rms} = 230 	ext{ V}$,$\omega = 300 	ext{ rad/s}$,and $C = 100 	ext{ pF} = 100 	imes 10^{-12} 	ext{ F}$.$I_{rms} = 230 	imes 300 	imes 100 	imes 10^{-12} 	ext{ A}$.$I_{rms} = 230 	imes 3 	imes 10^4 	imes 10^{-12} 	ext{ A}$.$I_{rms} = 690 	imes 10^{-8} 	ext{ A} = 6.9 	imes 10^{-6} 	ext{ A}$.

$A$ parallel plate capacitor having plates of radius $6 \text{ cm}$ has capacitance $100 \text{ pF}$. It is connected to a $230 \text{ V}$ a.c. supply with angular frequency $300 \text{ rad/s}$. The r.m.s. value of current is:

Similar Questions

Capacitive reactance of a capacitor in an $AC$ circuit is $3 \ k\Omega$. If this capacitor is connected to a new $AC$ source of double frequency,the capacitive reactance will become:

The reactance of a coil when used in the domestic $AC$ power supply $(220 \, V, 50 \, Hz)$ is $50 \, \Omega$. The inductance of the coil is nearly: (in $H$)

In a circuit containing a pure resistor connected to an $AC$ source,

$A$ choke coil is used as resistance in

An $AC$ source is connected to a capacitor $C$. Due to a decrease in its operating frequency:

Vedclass Test Series

Exam Paper Generator

Online Exam Module