A parallel plate capacitor in series with a r | vedclass

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

Question

(C) In a series $R-L-C$ circuit, the resonance frequency is given by $f_0 = \frac{1}{2\pi\sqrt{LC}}$.Given: $R = 100 \Omega$, $L = 20 	ext{ mH} = 20

Vedclass · Accepted Answer

$0.2 	ext{ mC}$

Vedclass · Answer

(C) In a series $R-L-C$ circuit, the resonance frequency is given by $f_0 = \frac{1}{2\pi\sqrt{LC}}$.Given: $R = 100 \Omega$, $L = 20 	ext{ mH} = 20 	imes 10^{-3} 	ext{ H}$, $f_0 = \frac{1250}{\pi} 	ext{ Hz}$.Substituting the values into the resonance formula:$\frac{1250}{\pi} = \frac{1}{2\pi\sqrt{20 	imes 10^{-3} 	imes C}}$$1250 = \frac{1}{2\sqrt{0.02 	imes C}}$$2500 = \frac{1}{\sqrt{0.02 	imes C}}$Squaring both sides:$6.25 	imes 10^6 = \frac{1}{0.02 	imes C}$$C = \frac{1}{0.02 	imes 6.25 	imes 10^6} = \frac{1}{0.125 	imes 10^6} = 8 	imes 10^{-6} 	ext{ F} = 8 \mu	ext{F}$.When charged by a $DC$ source of $V = 25 	ext{ V}$, the charge $Q$ stored is:$Q = C 	imes V = 8 	imes 10^{-6} 	ext{ F} 	imes 25 	ext{ V} = 200 	imes 10^{-6} 	ext{ C} = 0.2 	imes 10^{-3} 	ext{ C} = 0.2 	ext{ mC}$.

Similar Questions

Suppose the frequency of the source in the previous example can be varied.
$(a)$ What is the frequency of the source at which resonance occurs?
$(b)$ Calculate the impedance,the current,and the power dissipated at the resonant condition.

$A$ transmitting station releases waves of wavelength $960\, m$. $A$ capacitor of $2.56\, \mu F$ is used in the resonant circuit. The self-inductance of the coil necessary for resonance is $............ \times 10^{-8}\, H$.

For a series $LCR$ circuit with $L=2 \ H, C=18 \ \mu F$ and $R=10 \ \Omega$. What is the value of the $Q$ factor of this circuit?

If the reading of the voltmeter $V$ shown in the figure at resonance is $200 \, V$,then the quality factor of the circuit is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module