An alternating voltage, of angular frequency `omega` is induced in electric circuit consisting of inductance L and capacitance C, connected in parallel. Then across the inductance coil

Options

A. current is maximum when `omega^(2)=(1)/(LC)`
B. current is minimum when `omega^(2)=(1)/(LC)`
C. voltage is minimum when `omega^(2)=(1)/(LC)`
D. voltage is maximum when `omega^(2)=(1)/(LC)`

Answer

To solve the problem, we need to analyze the behavior of an electric circuit consisting of an inductor (L) and a capacitor (C) connected in parallel when an alternating voltage of angular frequency \( \omega \) is applied. We will determine the conditions under which the current and voltage reach their maximum and minimum values.

### Step-by-Step Solution:

1. **Understanding Resonance Condition**:
- In a parallel LC circuit, resonance occurs when the inductive reactance (\( X_L \)) equals the capacitive reactance (\( X_C \)).
- The inductive reactance is given by \( X_L = \omega L \) and the capacitive reactance is given by \( X_C = \frac{1}{\omega C} \).
- The resonance condition can be expressed as:
\[ X_L = X_C \implies \omega L = \frac{1}{\omega C} \]

2. **Finding the Resonant Frequency**:
- Rearranging the resonance condition gives:
\[ \omega^2 = \frac{1}{LC} \]
- Therefore, the resonant angular frequency \( \omega_0 \) is:
\[ \omega_0 = \frac{1}{\sqrt{LC}} \]

3. **Current Behavior at Resonance**:
- At resonance, the total current in the circuit is minimized because the inductive and capacitive currents cancel each other out.
- Thus, we can conclude:
\[ \text{Current is minimum when } \omega^2 = \frac{1}{LC} \]

4. **Voltage Behavior at Resonance**:
- In an inductor, the voltage leads the current by 90 degrees. Therefore, when the current is at its minimum (at resonance), the voltage across the inductor will be at its maximum.
- Hence, we can conclude:
\[ \text{Voltage is maximum when } \omega^2 = \frac{1}{LC} \]

5. **Final Conclusions**:
- From the analysis, we find that:
- The current is minimum when \( \omega^2 = \frac{1}{LC} \) (Option B).
- The voltage is maximum when \( \omega^2 = \frac{1}{LC} \) (Option D).

### Summary of Answers:
- **Correct Options**: B and D.