Physics Class 12 Part I…
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Lcr Series Circuit

🎓 Class 12 Physics CBSE Theory Ch 7 – Alternating Current ⏱ ~14 min
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Lcr Series Circuit

7.5 AC Voltage Applied to a Series LCR Circuit

Now we combine an inductor L, a capacitor C, and a resistor R in series across an AC source \(v = v_m \sin\omega t\) (Fig. 7.12). The same current \(i = i_m \sin(\omega t + \phi)\) flows through all three elements, but the voltages across them have different phase relations with this common current.

R L C v = vm sin ωt
Fig. 7.12: Series LCR circuit driven by AC source.

7.5.1 Phasor Diagram Solution

Take the current phasor \(\vec{I}\) along the +x axis. Then:

  • \(V_R = IR\), in phase with I → along +x.
  • \(V_L = IX_L\), leads I by 90° → along +y.
  • \(V_C = IX_C\), lags I by 90° → along -y.

Since VL and VC are anti-parallel, their net effect is \((V_L - V_C)\) along +y (assuming VL > VC). The source voltage phasor is the vector sum:

I (along VR) VL VC VL−VC V (source) φ
Fig. 7.13: Phasor diagram for series LCR. The source voltage leads the current by angle φ.

By Pythagoras' theorem:

\(v_m^2 = (i_m R)^2 + (i_m X_L - i_m X_C)^2\)
\(v_m = i_m\sqrt{R^2 + (X_L - X_C)^2}\)
Impedance: The total "effective resistance" of the LCR circuit:
\(Z = \sqrt{R^2 + (X_L - X_C)^2}\)
SI unit: ohm. Then \(i_m = v_m/Z\) and \(I = V/Z\). The phase angle \(\phi\) between source voltage and current is:
\(\tan\phi = \dfrac{X_L - X_C}{R}\)
\(\phi > 0\) → current lags voltage (inductive); \(\phi < 0\) → current leads (capacitive).

7.5.2 Impedance Triangle

Dividing every voltage by I (the common factor):

R XL − XC Z φ
Fig. 7.14: Impedance triangle. cos φ = R/Z; sin φ = (XL−XC)/Z.

7.5.3 Resonance

For fixed L, C and R, the impedance Z depends on the source frequency \(\omega\):

  • At low \(\omega\): \(X_C\) is huge, circuit is capacitive.
  • At high \(\omega\): \(X_L\) is huge, circuit is inductive.
  • At one special frequency \(\omega_0\): \(X_L = X_C\) and Z is minimum = R.

This condition is called resonance. Setting \(\omega L = 1/(\omega C)\):

\(\omega_0 = \dfrac{1}{\sqrt{LC}},\qquad f_0 = \dfrac{1}{2\pi\sqrt{LC}}\)

At resonance the current amplitude is largest: \(i_m^{max} = v_m/R\).

ω im ω0 vm/R 2Δω (bandwidth)
Fig. 7.16: Resonance curve - current amplitude peaks at ω = ω₀.

7.5.4 Sharpness of Resonance — Q-Factor

The sharpness of the resonance peak is characterised by the quality factor Q:

\(Q = \dfrac{\omega_0 L}{R} = \dfrac{1}{\omega_0 C R} = \dfrac{1}{R}\sqrt{\dfrac{L}{C}}\)

The full bandwidth (frequency span between the two half-power points, where I = Imax/√2) is:

\(2\Delta\omega = \dfrac{R}{L} = \dfrac{\omega_0}{Q}\)

Higher Q ⇒ sharper resonance ⇒ better selectivity (in radio tuning).

ConditionXL vs XCBehaviourPhase of i vs v
ω < ω₀XL < XCcapacitivei leads v
ω = ω₀XL = XCpurely resistive (Z = R)in phase (φ = 0)
ω > ω₀XL > XCinductivei lags v
Example 7.5 — Standard LCR problem

An AC source of 220 V (rms), 50 Hz is connected in series with R = 30 Ω, L = 80 mH and C = 60 μF. Find (a) impedance, (b) RMS current, (c) phase angle.

\(X_L = 2\pi \times 50 \times 0.08 = 25.13\) Ω

\(X_C = 1/(2\pi \times 50 \times 60\times 10^{-6}) = 53.05\) Ω

(a) \(Z = \sqrt{30^2 + (25.13-53.05)^2} = \sqrt{900+779.5} = \sqrt{1679.5} = 41.0\) Ω.

(b) \(I = V/Z = 220/41.0 = 5.37\) A.

(c) \(\tan\phi = (25.13 - 53.05)/30 = -0.93 \Rightarrow \phi = -43°\). Current LEADS voltage (capacitive circuit).

Example 7.6 — Find resonance and Q-factor

For the same L = 80 mH and C = 60 μF, find (a) resonance frequency f₀ and (b) Q-factor when R = 30 Ω.

(a) \(f_0 = 1/(2\pi\sqrt{LC}) = 1/(2\pi\sqrt{0.08\times 60\times 10^{-6}}) = 1/(2\pi\times 0.00219) = 72.6\) Hz.

(b) \(Q = (1/R)\sqrt{L/C} = (1/30)\sqrt{0.08/(60\times 10^{-6})} = (1/30)\sqrt{1333} = (1/30)(36.5) = 1.22\).

This is a low-Q (broad) resonance.

Simulation: Resonance Explorer

Adjust L, C and R. Read off the resonance frequency f₀, Q-factor and bandwidth.

Resonance frequency f₀72.6 Hz
Angular freq ω₀456 rad/s
Q-factor1.22
Bandwidth Δω375 rad/s
Peak current at resonance (for V=220V)7.33 A
Activity 7.3 — Tuning into a station

Set up an LC tank with a variable capacitor (gang capacitor from an old radio) and a few-turn coil.

Predict: what will happen as you slowly turn the capacitor dial while listening?
  1. Connect the tank to a sensitive AM receiver front-end (or use a smartphone AM signal-strength app near the coil).
  2. Slowly rotate the dial through its full range.

At one or two specific positions you hear stations come through loud and clear - the LC resonance frequency \(\omega_0 = 1/\sqrt{LC}\) matches the broadcast carrier frequency. Other positions give silence.

Competency-Based Questions L1L2L3L4L6

A series LCR circuit has L = 0.12 H, C = 480 nF, R = 23 Ω and is connected across a 230 V (rms) AC source whose frequency can be varied.

1. The resonance frequency of the circuit is closest to: L1

  • (a) 50 Hz
  • (b) 660 Hz
  • (c) 1.2 kHz
  • (d) 4.2 kHz
(b) 660 Hz. \(f_0 = 1/(2\pi\sqrt{0.12\times 480\times 10^{-9}}) = 1/(2\pi\times 2.40\times 10^{-4}) = 663\) Hz.

2. At resonance the impedance of this circuit equals which quantity? Justify briefly. L2

Z = R = 23 Ω. At resonance XL = XC, so their difference is zero and \(Z=\sqrt{R^2+0}\) = R.

3. Find the maximum RMS current at resonance. L3

\(I_{max} = V/R = 230/23 = 10.0\) A.

4. The same circuit is now driven at 200 Hz (well below resonance). Is the current leading or lagging the voltage? Explain. L4

At 200 Hz: XL = 2π·200·0.12 ≈ 151 Ω; XC = 1/(2π·200·480·10⁻⁹) ≈ 1658 Ω. XC >> XL, circuit is capacitive, so current LEADS voltage. \(\tan\phi = (151-1658)/23 = -65.5\) ⇒ φ ≈ −89°.

5. Design a circuit to act as a sharp 1 MHz filter. State your L, C, R choices and the resulting Q. L6

Pick L = 100 μH, C = 253 pF: \(f_0 = 1/(2\pi\sqrt{LC}) \approx 1\) MHz. Choose R = 1 Ω for a high-Q peak: \(Q = (1/R)\sqrt{L/C} = \sqrt{100\times 10^{-6}/253\times 10^{-12}} ≈ 628\). Bandwidth ≈ 1.6 kHz - sharp enough to pick out one AM broadcaster.

Assertion-Reason Questions

Options: (A) Both true, R explains A. (B) Both true, R does not explain A. (C) A true, R false. (D) A false, R true.

Assertion: At resonance the current in a series LCR circuit is maximum.

Reason: At resonance the inductive and capacitive reactances cancel, leaving impedance = R alone.

(A). Both true and R correctly explains A.

Assertion: A high Q-factor means a broad resonance peak.

Reason: Bandwidth = ω₀ / Q.

(D). A is false (high Q → SHARP, NARROW peak). R is true.

Assertion: The voltage across L can be larger than the source voltage at resonance.

Reason: At resonance VL = IXL = Q × Vsource, which exceeds Vsource when Q > 1.

(A). Both true; R explains A. This "voltage magnification" by Q is a key feature of LCR resonance.

Frequently Asked Questions - Lcr Series Circuit

What is the main concept covered in Lcr Series Circuit?
In NCERT Class 12 Physics Chapter 7 (Alternating Current), "Lcr Series Circuit" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Lcr Series Circuit useful in real-life applications?
Real-life applications of "Lcr Series Circuit" from NCERT Class 12 Physics Chapter 7 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Lcr Series Circuit?
Key formulas in "Lcr Series Circuit" (NCERT Class 12 Physics Chapter 7 Alternating Current) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 7?
NCERT Class 12 Physics Chapter 7 (Alternating Current) is structured so each part builds on the previous one. "Lcr Series Circuit" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Lcr Series Circuit?
CBSE board questions from "Lcr Series Circuit" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Lcr Series Circuit" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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