In an LCR series circuit,C = 2 mu F,L = 1 mH | vedclass

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

Question

(A) In an $LCR$ series circuit,the current is maximum at resonance,where the inductive reactance equals the capacitive reactance $(X_L = X_C)$.At reso

Vedclass · Accepted Answer

$1 : 1$

Vedclass · Answer

(A) In an $LCR$ series circuit,the current is maximum at resonance,where the inductive reactance equals the capacitive reactance $(X_L = X_C)$.At resonance,the impedance $Z = R$,so the maximum current is $I_{\max} = \frac{V}{R}$.The energy stored in the inductor is $W_L = \frac{1}{2} L I_{\max}^2$.The energy stored in the capacitor is $W_C = \frac{1}{2} C V_C^2$,where $V_C = I_{\max} X_C$.Since $X_C = \frac{1}{\omega C}$ and $X_L = \omega L$,at resonance $\omega = \frac{1}{\sqrt{LC}}$.Thus,$X_C = X_L = \sqrt{\frac{L}{C}}$.$W_C = \frac{1}{2} C (I_{\max} X_C)^2 = \frac{1}{2} C I_{\max}^2 (\frac{L}{C}) = \frac{1}{2} L I_{\max}^2$.Wait,let's re-evaluate: $W_C = \frac{1}{2} C V_C^2 = \frac{1}{2} C (I_{\max} \cdot \frac{1}{\omega C})^2 = \frac{1}{2} C I_{\max}^2 \cdot \frac{1}{\omega^2 C^2} = \frac{I_{\max}^2}{2 \omega^2 C}$.Substituting $\omega^2 = \frac{1}{LC}$,we get $W_C = \frac{I_{\max}^2}{2 (1/LC) C} = \frac{1}{2} L I_{\max}^2$.Therefore,$W_C = W_L$,so the ratio is $1:1$.

In an $LCR$ series circuit,$C = 2 \mu F$,$L = 1 \ mH$,and $R = 10 \ \Omega$. When the current in the circuit is maximum,what is the ratio of the energy stored in the capacitor to the energy stored in the inductor?

Similar Questions

The quality factor of an $LCR$ circuit having resistance $(R)$ and inductance $(L)$ at resonance frequency $(\omega)$ is given by:

For an $R-L-C$ circuit,driven with voltage of amplitude $V_m$ and frequency $\omega_0 = \frac{1}{\sqrt{LC}}$,the current exhibits resonance. The quality factor $Q$ is

An inductor of $0.5 \text{ mH}$, a capacitor of $20 \text{ } \mu\text{F}$ and resistance $20 \text{ } \Omega$ are connected in series with a $220 \text{ V}$ ac source. If the current is in phase with the emf, the amplitude of current of the circuit is $[x]^{1/2} \text{ A}$. The value of $x$ is

For an $L-C-R$ $AC$ circuit,the resonance frequency is $5000 \ Hz$ and the frequencies at half-power points are $4950 \ Hz$ and $5050 \ Hz$. What will be the $Q$-factor?

In an $L-C-R$ series $AC$ circuit,$L = 9 \ H$,$R = 10 \ \Omega$,and $C = 100 \ \mu F$. The $Q$-factor of the circuit is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module