In a series LCR circuit, C = 2 mu F, L = 5 te | vedclass

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

Question

(A) In a series $LCR$ circuit, maximum current flows through the circuit at resonance.At resonance, the inductive reactance equals the capacitive reac

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(A) In a series $LCR$ circuit, maximum current flows through the circuit at resonance.At resonance, the inductive reactance equals the capacitive reactance, i.e., $X_L = X_C$.The energy stored in the inductor is given by $U_L = \frac{1}{2} L I_{max}^2$.The energy stored in the capacitor is given by $U_C = \frac{1}{2} C V_C^2$, where $V_C = I_{max} X_C$.Since $X_L = X_C = \omega L = \frac{1}{\omega C}$, we have $V_C = I_{max} X_L = I_{max} \omega L$.Substituting this into the energy formula: $U_C = \frac{1}{2} C (I_{max} \omega L)^2 = \frac{1}{2} C I_{max}^2 \omega^2 L^2$.Since $\omega^2 = \frac{1}{LC}$, we get $U_C = \frac{1}{2} C I_{max}^2 (\frac{1}{LC}) L^2 = \frac{1}{2} L I_{max}^2$.Thus, at resonance, $U_L = U_C$.However, the question asks for the ratio based on the given values. Let us re-evaluate: $U_L = \frac{1}{2} L I_{max}^2$ and $U_C = \frac{1}{2} C V_C^2$. At resonance, $V_C = I_{max} X_C$. Since $X_C = \frac{1}{\omega C}$ and $\omega = \frac{1}{\sqrt{LC}}$, $X_C = \sqrt{\frac{L}{C}}$.$U_C = \frac{1}{2} C (I_{max} \sqrt{\frac{L}{C}})^2 = \frac{1}{2} C I_{max}^2 (\frac{L}{C}) = \frac{1}{2} L I_{max}^2$.Wait, the ratio $U_L/U_C$ at resonance is $1:1$. Given the options, there might be a misunderstanding of the state. If the question implies the ratio of maximum energy stored in the inductor to the maximum energy stored in the capacitor (not necessarily at the same time), then $U_{L,max} = \frac{1}{2} L I_{max}^2$ and $U_{C,max} = \frac{1}{2} C V_{max}^2$. With $V_{max} = I_{max} Z$, this is not constant. Re-reading: The ratio of energy stored in the inductor to that in the capacitor when maximum current flows is $1:1$. Given the options, let's check $L/C$ ratio: $L/C = (5 	imes 10^{-3}) / (2 	imes 10^{-6}) = 2500$. None match. If the question implies $U_L/U_C = (\frac{1}{2} L I^2) / (\frac{1}{2} Q^2/C)$, at resonance $Q = I/\omega$. Ratio $= L / (1/(\omega^2 C)) = L \omega^2 C = 1$. Given the options, there is a discrepancy in the question premise.

In a series $LCR$ circuit, $C = 2 \mu F$, $L = 5 \text{ mH}$, and $R = 5 \Omega$. What is the ratio of the energy stored in the inductor to that in the capacitor when the maximum current flows through the circuit (in $:$)?

Similar Questions

An $LCR$ series circuit of capacitance $62.5 \, nF$ and resistance of $50 \, \Omega$ is connected to an $A.C.$ source of frequency $2.0 \, kHz$. For the maximum value of the amplitude of current in the circuit,the value of inductance is $.......... \, mH$. (take $\pi^2 = 10$)

At resonance,the value of current in a series $L-C-R$ circuit is: (Symbols have their usual meanings.)

What will be the reading of the voltmeter in the given circuit (in $V$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module