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

Concept: Resonance in AC Circuits — the phase angle $\phi$ depends on whether the inductive reactance $X_L = \omega L$ or capacitive reactance $X_C = 1/(\omega C)$ dominates.

Reasoning:

1. For an LCR series circuit, $\tan \phi = \frac{X_L - X_C}{R} = \frac{\omega L - 1/(\omega C)}{R}$.
2. At very low frequencies, $X_C \gg X_L$, so $X_L - X_C < 0$ → $\phi$ is negative (current leads voltage).
3. At resonance ($\omega = \omega_0 = 1/\sqrt{LC}$), $X_L = X_C$ → $\phi = 0$. …

The behaviour of an LCR series circuit is governed by one central idea: resonance. At any frequency, the total opposition to current is the impedance $Z$, which combines resistance $R$, inductive reactance $X_L = \omega L$, and capacitive reactance $X_C = \frac{1}{\omega C}$. The phase angle $\phi$ tells us whether the circuit looks more like a capacitor (current leads) or an inductor (current lags).

The formula for $\phi$ is:

The sign of $\phi$ depends entirely on whether $X_L > X_C$ (positive $\phi$, voltage leads) or $X_L < X_C$ (negative $\phi$, current leads). At resonance, $X_L = X_C$, so $\phi = 0$.

Now let's track what happens as frequency sweeps from very low to very high.

1. At very low frequencies ($\omega \to 0$)

   • $X_L = \omega L \to 0$ (inductor acts like a short)
   • $X_C = \frac{1}{\omega C} \to \infty$ (capacitor blocks current)
   • So $X_L - X_C \approx -\infty$, making $\tan \phi \to -\infty$
   • This means $\phi \to -\frac{\pi}{2}$ (or $-90^\circ$)
   • Current leads voltage — the circuit is predominantly capacitive.

2. As frequency increases (but still below resonance)

   • $X_L$ grows linearly, $X_C$ falls hyperbolically
   • The difference $X_L - X_C$ becomes less negative
   • $\phi$ rises from $-\frac{\pi}{2}$ toward zero
   • The circuit becomes less capacitive, more resistive.

3. At resonance ($\omega = \omega_0 = \frac{1}{\sqrt{LC}}$)

   • $X_L = X_C$, so $\tan \phi = 0$
   • $\phi = 0$ — voltage and current are in phase
   • The circuit behaves as a pure resistor; impedance is minimum ($Z = R$).

4. Above resonance ($\omega > \omega_0$)

   • $X_L$ now exceeds $X_C$
   • $X_L - X_C$ becomes positive, so $\tan \phi > 0$
   • $\phi$ becomes positive and increases toward $+\frac{\pi}{2}$
   • Voltage leads current — the circuit is predominantly inductive.

5. At very high frequencies ($\omega \to \infty$)

   • $X_L \to \infty$ (inductor blocks), $X_C \to 0$ (capacitor shorts)
   • $X_L - X_C \to +\infty$, so $\tan \phi \to +\infty$
   • $\phi \to +\frac{\pi}{2}$ (or $+90^\circ$)
   • Voltage leads current by a quarter cycle. …

Method: Tracking How the Phase Angle Changes Across a Range of Frequencies in a Series LCR Circuit

This method applies to any question asking how a series LCR circuit's phase angle, impedance character, or "which element dominates" changes as frequency is swept from very low to very high.

Steps

Step 1: Write the general phase-angle formula

This single formula governs the phase angle at every frequency — track the SIGN of the numerator as $\omega$ changes.

Step 2: Evaluate the two reactances at the extremes

As $\omega \to 0$: $X_L = \omega L \to 0$ (inductor looks like a short) while $X_C = 1/(\omega C) \to \infty$ (capacitor blocks). As $\omega \to \infty$: the opposite happens — $X_L \to \infty$, $X_C \to 0$. This tells you the circuit is capacitive-dominated at low frequency and inductive-dominated at high frequency.

Step 3: Identify the crossover point — resonance …