Question

In series LCR circuit C = 2 uF, L=1 mH and R= 10 \(\Omega\). What is the ratio of energies stored in the capacitor and inductor when maximum current flows through circuit?

• A. \(0.1:1\)
• B. \(0.4:1\)
• C. \(0.2:1\)
• D. \(1:1\)

Hint:

Maximum current occurs at resonance where \(X_L = X_C\), and then the energy in the capacitor equals the energy in the inductor.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In a series LCR circuit the current is largest when the circuit is at resonance. This is the condition where the inductive reactance equals the capacitive reactance.

Step 2: Key Formula or Approach:
At resonance \(X_L = X_C\), that is \(\omega L = \frac{1}{\omega C}\), which gives \(\omega^2 = \frac{1}{LC}\).
The energy in the inductor is \(U_L = \frac{1}{2}LI^2\). The energy in the capacitor is \(U_C = \frac{1}{2}CV_C^2\), where \(V_C = I X_C\).

Step 3: Detailed Explanation:
Put \(V_C = I X_C = \dfrac{I}{\omega C}\) into the capacitor energy:
\[ U_C = \frac{1}{2} C \frac{I^2}{\omega^2 C^2} = \frac{1}{2}\frac{I^2}{\omega^2 C} \]
Now use \(\omega^2 = \frac{1}{LC}\), so \(\frac{1}{\omega^2 C} = L\):
\[ U_C = \frac{1}{2} L I^2 = U_L \]
So the two energies are equal and
\[ U_C : U_L = 1 : 1 \]
The values \(C = 2\ \mu\text{F}\), \(L = 1\ \text{mH}\) and \(R = 10\ \Omega\) are not needed, because the result holds for any \(L\), \(C\) and \(R\). The resistance only fixes how large the current is, not the ratio.
Options (A), (B) and (C) give ratios different from 1, and none of them can come from the resonance condition.

Final Answer:
The ratio is 1:1, option (D). \[ \boxed{1:1} \]