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#AC circuits

30 public questions tagged with this topic.

Resonance in LCR Circuit and Q-Factor

In an LCR series circuit, if the frequency is below the resonant frequency, which component dominates the circuit's beha

**At resonance** V_L = I X_L = I X_C = V_C, may be larger than source voltage Q times, Q-factor = ω₀ L/R =1/(ω₀ C R)= V_L/V = V_C/V, measures sharpness, higher Q sharper resonance, bandwidth Δω = R/L = ω₀/Q, resonant frequency independent of R. Below resonance, ω is low, making X_C = (1/ω C) larger than X_L = ω L . Thus, the capacitive reactance dominates, and the circuit behaves as predominantly capacitive, with current leading voltage. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation

Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor

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AC Through Capacitor - Capacitive Reactance

A \( 130 \, \text{V} \) (rms) AC source supplies a \( 65 \, \Omega \) resistor. What is the average power consumed?

**AC through capacitor** voltage lags current by 90°, Q= C V, I= dQ/dt = C dV/dt, V(t)=V_peak sin ωt, I(t)=I_peak sin(ωt+90°), average power zero because energy stored in electric field ½ C V² returned each cycle, capacitor blocks DC but passes AC, X_C decreases with f. RMS current: I = (V/R) = (130/65) = 2 A . Average power: P = I² R = 2² × 65 = 260 W . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives 260 W, consistent with phasor analysis and resonance condition X_L = X_C.

Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance

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Power in AC Circuits - Power Factor and Wattless Current

A series LCR circuit has \( R = 20 \, \Omega \), \( X_L = 30 \, \Omega \), \( X_C = 50 \, \Omega \). What is the power f

**Power factor** cos φ = R/Z, 0≤cos φ≤1, at resonance cos φ=1 maximum power, pure L or C cos φ=0 zero average power, current wattless because energy stored returned. Instantaneous power when current maximum in pure L: I=I_peak, V=0 because V leads 90°, so P= V I =0 at that instant, average zero. Z = √(R² + (X_L - X_C)²) = √(20² + (30 - 50)²) = √(400 + 400) = √(800) ≈ 28.28 Ω . Power factor: cos Φ = (R/Z) = (20/28.28) ≈ 0.707 . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P

Ref: NCERT > Physics Book > Alternating Currents > Power in AC Circuits - Power Factor and Wattless Current

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AC Fundamentals - RMS, Average and Peak Values

A \( 25 \, \Omega \) resistor is connected to a \( 125 \, \text{V} \) (rms) AC source. What is the rms current?

**Average power** in resistor P_avg = V_rms I_rms = V_rms²/R = I_rms² R, for 254.6 V peak, V_rms =180 V, R=90 Ω, P=180²/90=360 W, for 226.3 V peak, V_rms=160 V, R=80 Ω, P=320 W, illustrating rms use for power. RMS current: I = (V/R) . Given: V = 125 V , R = 25 Ω . I = (125/25) = 5 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives 5 A, consistent with phasor analysis and resonance condition X_L = X_C.

Ref: NCERT > Physics Book > Alternating Currents > AC Fundamentals - RMS, Average and Peak Values

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AC Fundamentals - RMS, Average and Peak Values

A \( 210 \, \text{V} \) (rms) source supplies a \( 105 \, \Omega \) resistor. What is the peak current?

**Average power** in resistor P_avg = V_rms I_rms = V_rms²/R = I_rms² R, for 254.6 V peak, V_rms =180 V, R=90 Ω, P=180²/90=360 W, for 226.3 V peak, V_rms=160 V, R=80 Ω, P=320 W, illustrating rms use for power. RMS current: I = (V/R) = (210/105) = 2 A . Peak current: i_m = √(2) I = 1.414 × 2 ≈ 2.828 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives 2.828 A, consistent with phasor analysis and resonance condition X_L = X_C.

Ref: NCERT > Physics Book > Alternating Currents > AC Fundamentals - RMS, Average and Peak Values

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AC Fundamentals - RMS, Average and Peak Values

A \( 160 \, \text{V} \) (rms) AC source supplies a \( 80 \, \Omega \) resistor. What is the average power consumed?

**RMS value** I_rms = I_peak/√2, V_rms = V_peak/√2 for sinusoidal AC, significance rms gives equivalent DC value producing same heating power P = I_rms² R, average power over cycle, instruments measure rms, average over full cycle zero, half-cycle average 2 I_peak/π, peak = √2 rms. RMS current: I = (V/R) = (160/80) = 2 A . Average power: P = I² R = 2² × 80 = 320 W . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives 320 W, consistent with phasor analysis and resonance condition X_L = X_C.

Ref: NCERT > Physics Book > Alternating Currents > AC Fundamentals - RMS, Average and Peak Values

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Resonance in LCR Circuit and Q-Factor

In a series LCR circuit, under what condition does the circuit behave as if it has only resistance?

**Resonance in LCR** occurs when X_L = X_C, ω₀ =1/√(LC), f₀=1/(2π√(LC)), impedance Z=R minimum, current maximum I₀=V/R, circuit behaves as if only resistance because inductive and capacitive voltages equal opposite cancel, net reactance zero, phase φ=0°, power factor 1, current in phase with voltage. In a series LCR circuit, the circuit behaves as purely resistive at resonance, where the inductive reactance equals the capacitive reactance ( X_L = X_C ). This cancels the reactive components, leaving only the resistance to determine the impedance. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L -

Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor

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LCR Series Circuit - Impedance and Phasor Diagram

What is the behavior of the impedance in a series LCR circuit at very high frequencies?

**Impedance behavior at high frequencies** X_L=ωL dominates ∝ f, X_C=1/ωC →0, so Z≈√(R²+X_L²)≈X_L large, current small, circuit inductive, φ→90°, at low frequencies X_C large, Z≈X_C, capacitive, φ→-90°, at intermediate resonance Z minimal =R. At very high frequencies, X_L = ω L becomes very large, while X_C = (1/ω C) becomes very small. The impedance Z = √(R² + (X_L - X_C)²) is dominated by X_L , making the circuit behave as predominantly inductive. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives It behaves as

Ref: NCERT > Physics Book > Alternating Currents > LCR Series Circuit - Impedance and Phasor Diagram

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PHYSICS

A 12 μF capacitor is connected to a 220 V, 50 Hz AC source. What is the rms current?

Given: A 12 μF capacitor is connected to a 220 V, 50 Hz AC source. What is the rms current? Formula: X_C = 1/omega C, omega = 2π × 50 = 314 rad/s. Substitution & Calculation: C = 12 × 10⁻⁶F . X_C = frac1314 × 12 × 10⁻⁶approx 265.3 Ω . RMS current: I = V/X_C = 220/265.3 approx 0.83 A . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

What happens to the current in a purely capacitive AC circuit when the frequency of the source increases?

In a purely capacitive circuit, the capacitive reactance ( X_C = 1/omega C ) decreases as the frequency ( f, where omega = 2Ï€ f ) increases. Since current is inversely proportional to reactance ( I = V/X_C ), the current increases.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Current Electricity, Topic: Resistivity, temperature dependence and circuit analysis.

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