Resonance in the LCR Circuit

The Swing Analogy

Resonance is common to all systems that tend to oscillate at a natural frequency: drive such a system near that frequency and the oscillation amplitude becomes large. NCERT's example — a child on a swing. The swing has its own back-and-forth frequency; pulls timed to match it build a large amplitude from gentle pushes.

The series LCR circuit is exactly such a system. Its current amplitude

$$i_m = \frac{v_m}{\sqrt{R^2 + (X_C - X_L)^2}}$$

depends on $\omega$ through $X_L = \omega L$ (rising) and $X_C = 1/\omega C$ (falling). At one special frequency they cross.

The Resonant Frequency

Set $X_C = X_L$:

$$\frac{1}{\omega_0 C} = \omega_0 L \quad\Rightarrow\quad \boxed{\omega_0 = \frac{1}{\sqrt{LC}}} \qquad \left(\nu_0 = \frac{1}{2\pi\sqrt{LC}}\right)$$

At $\omega_0$:

• The impedance is minimum: $Z = \sqrt{R^2 + 0} = R$.
• The current amplitude is maximum: $i_m = v_m/R$.
• $\phi = 0$: current and voltage in phase; the circuit behaves as purely resistive.
• $V_L$ and $V_C$ are individually large but cancel exactly — the whole source voltage appears across R.

NCERT's plotted case: L = 1.00 mH, C = 1.00 nF gives $\omega_0 = \frac{1}{\sqrt{10^{-3} \times 10^{-9}}} = 1.00 \times 10^6$ rad/s; at resonance the peak current for R = 100 $\Omega$ is exactly twice that for R = 200 $\Omega$ — smaller R, taller and sharper peak.

Key Point (NCERT, emphatic): resonance is exhibited only if both L and C are present — only then can $V_L$ and $V_C$ (out of phase) cancel. There is no resonance in RL or RC circuits.

Tuning, and Other Uses of a Peak

Radio/TV tuning is resonance at work. The antenna feeds the tuning circuit signals from many stations — many driving frequencies at once. Turning the tuner varies a capacitor, changing the circuit's $\nu_0$ until it matches your station's frequency; that signal alone drives a large current. Selection by resonance.

The airport metal detector (NCERT Example 7.10) is the same physics: the detector coil is part of a circuit tuned near resonance. Metal you carry changes the effective inductance, shifts the circuit off resonance, changes the current — and the change triggers the alarm.

[JEE Tip] At resonance the element voltages are $V_R = V$ (full source), and $V_L = V_C = \frac{X_L}{R}V$ — which can be many times the source voltage (voltage magnification). Computing $V_L$ at resonance and being unafraid when it dwarfs the source is a JEE hallmark.

[NEET Important] Three resonance facts asked endlessly: $Z_{min} = R$, $I_{max} = V/R$, $\cos\phi = 1$. And the frequency: $\nu_0 = \frac{1}{2\pi\sqrt{LC}}$ — sharpen your $\sqrt{LC}$ arithmetic.

Solved Examples

Example 1: NCERT's plotted circuit [JEE Numerical]

Find the resonant angular frequency and frequency for L = 1.00 mH and C = 1.00 nF.

Solution:

1. $LC = 10^{-3} \times 10^{-9} = 10^{-12}$, so $\sqrt{LC} = 10^{-6}$ s.
2. $\omega_0 = 1/\sqrt{LC} = 1.00 \times 10^6$ rad/s.
3. $\nu_0 = \omega_0/2\pi \approx 159$ kHz — NCERT's Fig. 7.14 case exactly.

Example 2: Resonance of the 283 V circuit (NCERT Example 7.9 setup) [JEE Numerical]

For L = 25.48 mH and C = 796 $\mu$F, find the resonant frequency; with $v_m$ = 283 V and R = 3 $\Omega$, find the peak current at resonance.

Solution:

1. $LC = 25.48 \times 10^{-3} \times 796 \times 10^{-6} = 2.028 \times 10^{-5}$; $\sqrt{LC} = 4.5 \times 10^{-3}$ s.
2. $\omega_0 = 222$ rad/s ($\nu_0 \approx 35.4$ Hz).
3. At resonance $Z = R$: $i_m = v_m/R = 283/3 = 94.3$ A — compare 56.6 A off resonance (Section 6).

Example 3: Tuning a radio [JEE Numerical]

A tuning circuit has L = 200 $\mu$H. What capacitance tunes it to an 800 kHz station?

Solution:

1. Condition: $\omega_0^2 = 1/LC$, so $C = \frac{1}{\omega_0^2 L}$.
2. $\omega_0 = 2\pi \times 8 \times 10^5 = 5.03 \times 10^6$ rad/s; $\omega_0^2 = 2.53 \times 10^{13}$.
3. $C = \frac{1}{2.53 \times 10^{13} \times 2 \times 10^{-4}} \approx 2.0 \times 10^{-10}$ F $= 0.2$ nF.
4. Turning the tuning knob is sweeping exactly this C.

Example 4: Voltage magnification at resonance [JEE Numerical]

At resonance, a series circuit (R = 10 $\Omega$, $X_L = X_C = 100\ \Omega$) is driven by a 220 V source. Find the current and the rms voltage across the inductor.

Solution:

1. $Z = R$: $I = 220/10 = 22$ A.
2. $V_L = IX_L = 22 \times 100 = 2200$ V — ten times the source!
3. $V_C = 2200$ V too, in exact anti-phase: they cancel, leaving 220 V across R. Large internal voltages, perfectly balanced books.

Example 5: Scaling the resonant frequency [NEET Numerical]

The inductance of a resonant circuit is quadrupled (C unchanged). What happens to $\nu_0$?

Solution:

1. $\nu_0 \propto 1/\sqrt{LC} \propto 1/\sqrt{L}$.
2. $L \to 4L$ gives $\nu_0 \to \nu_0/2$.
3. Answer: the resonant frequency halves. (Quadrupling C would do the same.)

Example 6: Power factor and power at resonance [NEET Numerical]

At resonance, a series LCR circuit (R = 5 $\Omega$) carries an rms current of 10 A. Find the power factor and the power drawn.

Solution:

1. At resonance $\phi = 0$: power factor $\cos\phi = 1$ (maximum).
2. $P = I^2R = 100 \times 5 = 500$ W.
3. All of it dissipates in R — L and C are merely exchanging energy with each other.

Example 7: Why no resonance in RL or RC?

Explain NCERT's insistence that resonance needs both L and C.

Solution:

1. Resonance is the cancellation of $V_L$ by $V_C$ — possible only because they are 180 degrees apart in phase.
2. With one reactance absent, nothing can cancel the other: Z = $\sqrt{R^2 + X^2}$ never falls to R at any finite frequency.
3. No minimum in Z, no current peak — no resonance in RL or RC circuits.

Example 8: How a radio selects one station

The antenna delivers signals at many frequencies. Why does only one station play?

Solution:

1. Each signal drives the tuning circuit at its own frequency; the circuit responds strongly only near its resonant frequency.
2. Tuning varies C until $\nu_0 = \frac{1}{2\pi\sqrt{LC}}$ matches the desired station.
3. That station's current dwarfs the rest — amplitude selection by resonance.

Example 9: The airport metal detector (NCERT Example 7.10)

Why does walking through a metal detector with coins set it off?

Solution:

1. The detector's gateway coil is part of a circuit tuned near resonance with large current.
2. Metal passing through changes the coil's effective inductance, shifting $\nu_0$ — the circuit moves off resonance and the current changes significantly.
3. Electronics sense the current change and sound the alarm. Resonance turned into security.

Example 10: Reading the resonance curve

NCERT's $i_m$-vs-$\omega$ plot shows two curves for R = 100 $\Omega$ and 200 $\Omega$ (same L, C, $v_m$ = 100 V). Compare their peaks and positions.

Solution:

1. Position: both peak at the same $\omega_0 = 1/\sqrt{LC}$ — R does not move the resonance.
2. Height: peak current $= v_m/R$: 1.0 A for 100 $\Omega$, 0.5 A for 200 $\Omega$ — half.
3. Smaller R also makes the peak sharper — the circuit is more selective (the idea behind good tuners).