How does the sign of the phase angle `phi`, by which the supply voltage leads the current in an LCR serices circuit, change as the supply frequency is gradually increased from very low to very high values ?

Explanation

To analyze how the sign of the phase angle \( \phi \) changes as the supply frequency in an LCR series circuit is gradually increased from very low to very high values, we can follow these steps:

### Step 1: Understand the Phase Angle Formula
The phase angle \( \phi \) in an LCR circuit is given by the equation:
\[ \tan \phi = \frac{X_L - X_C}{R} \]
where:
- \( X_L = 2 \pi f L \) (inductive reactance)
- \( X_C = \frac{1}{2 \pi f C} \) (capacitive reactance)
- \( R \) is the resistance.

### Step 2: Analyze at Very Low Frequency
When the frequency \( f \) is very low:
- \( X_L = 2 \pi f L \) becomes a small value.
- \( X_C = \frac{1}{2 \pi f C} \) becomes a very large value because it is inversely proportional to \( f \).

Thus, we have:
\[ \tan \phi = \frac{X_L - X_C}{R} \approx \frac{\text{small} - \text{large}}{R} < 0 \]
This implies that \( \phi < 0 \) (the voltage lags behind the current).

### Step 3: Analyze at Resonant Frequency
At the resonant frequency \( f_r \):
- At resonance, \( X_L = X_C \), which leads to:
\[ \tan \phi = \frac{X_L - X_C}{R} = 0 \]
Thus, \( \phi = 0 \) (the voltage and current are in phase).

### Step 4: Analyze at Very High Frequency
When the frequency \( f \) is very high:
- \( X_L = 2 \pi f L \) becomes a very large value.
- \( X_C = \frac{1}{2 \pi f C} \) becomes a very small value.

Thus, we have:
\[ \tan \phi = \frac{X_L - X_C}{R} \approx \frac{\text{large} - \text{small}}{R} > 0 \]
This implies that \( \phi > 0 \) (the voltage leads the current).

### Conclusion
As the supply frequency increases from very low to very high:
- At very low frequency: \( \phi < 0 \) (voltage lags current)
- At resonance: \( \phi = 0 \) (voltage and current are in phase)
- At very high frequency: \( \phi > 0 \) (voltage leads current)