An L.C, R series circuit is connected to ac, source. At resonance, the applied voltage and the current flowing through the circuit will have a phase difference of

Options

A. `pi//4`
B. zero
C. pi
D. `pi//2`

Answer

To solve the problem of finding the phase difference between the applied voltage and the current flowing through an LCR series circuit at resonance, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- An LCR series circuit consists of an inductor (L), capacitor (C), and resistor (R) connected in series to an AC voltage source.
- At resonance, the inductive reactance (X_L) and capacitive reactance (X_C) are equal.

2. **Condition at Resonance**:
- At resonance, the condition is given by:
\[ X_L = X_C \]
- This implies that the current through the inductor (I_L) and the current through the capacitor (I_C) are equal in magnitude but opposite in phase.

3. **Current in the Circuit**:
- The current through the resistor (I_R) is in phase with the applied voltage (V).
- Since I_L and I_C are equal and opposite, they effectively cancel each other out at resonance.

4. **Net Current**:
- The net current (I) in the circuit is the same as the current through the resistor (I_R) because the effects of I_L and I_C cancel each other out.
- Therefore, the net current is in phase with the applied voltage.

5. **Phase Difference**:
- Since the net current (I) is in phase with the applied voltage (V), the phase difference between the applied voltage and the current flowing through the circuit at resonance is:
\[ \text{Phase Difference} = 0^\circ \]

### Conclusion:
At resonance in an LCR series circuit, the phase difference between the applied voltage and the current flowing through the circuit is **0 degrees**.