Question

In a discharging series L-C-R circuit, \(C=2\ \mu \text{F}\), \(L=1\text{ mH}\) and \(R=10\ \Omega\). The ratio of the energies stored in the inductor and in the capacitor, when the maximum current flows in the circuit, is

• A. \(2:1\)
• B. \(3:1\)
• C. \(3:2\)
• D. \(5:1\)

Hint:

At maximum current the rate of change of current is zero, so the capacitor voltage equals the voltage drop across R.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
In a series L-C-R circuit the loop equation is \(L\dfrac{dI}{dt}+IR+\dfrac{q}{C}=0\). When the current is maximum, \(\dfrac{dI}{dt}=0\).

Step 2: Capacitor Voltage
This gives \(\dfrac{q}{C}=IR\) in magnitude.

Step 3: Energies
Inductor: \(U_L=\dfrac{1}{2}LI^{2}\)
Capacitor: \(U_C=\dfrac{1}{2}C\left(IR\right)^{2}\)

Step 4: Calculation
\(\dfrac{U_L}{U_C}=\dfrac{L}{CR^{2}}=\dfrac{10^{-3}}{2\times10^{-6}\times100}=5\)

Step 5: Compare the options
The ratio is \(5:1\), the last option.

Final Answer:
The ratio of the energy in the inductor to that in the capacitor is 5 : 1. \[ \boxed{\text{(D) }5:1} \]