An L-C-R circuit is connected to a source of A.C. current. At resonance, the phase difference between the applied voltage and the current in the circuit, is

Options

A. `pi`
B. Zero
C. `(pi)/(4)`
D. `(pi)/(2)`

Answer

To solve the question regarding the phase difference between the applied voltage and the current in an L-C-R circuit at resonance, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Resonance in an L-C-R Circuit**:
- In an L-C-R circuit, resonance occurs when the inductive reactance (\(X_L\)) is equal to the capacitive reactance (\(X_C\)). This condition can be expressed as:
\[ X_L = X_C \]

2. **Impedance at Resonance**:
- The total impedance (\(Z\)) of the circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
- At resonance, since \(X_L = X_C\), we have:
\[ Z = R \]

3. **Phase Angle Calculation**:
- The phase angle (\(\phi\)) between the voltage and the current is given by:
\[ \tan(\phi) = \frac{X_L - X_C}{R} \]
- Since \(X_L - X_C = 0\) at resonance, we can substitute this into the equation:
\[ \tan(\phi) = \frac{0}{R} = 0 \]

4. **Determining the Phase Difference**:
- If \(\tan(\phi) = 0\), then the phase angle \(\phi\) is:
\[ \phi = 0^\circ \]
- This means that at resonance, the voltage and current are in phase.

5. **Conclusion**:
- Therefore, the phase difference between the applied voltage and the current in the circuit at resonance is \(0^\circ\).

### Final Answer:
At resonance, the phase difference between the applied voltage and the current in the circuit is **0 degrees**.

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