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#LCR circuit

47 public questions tagged with this topic.

What is the effect on the impedance of a series LCR circuit when the frequency is increased beyond the resonant frequenc

**Transformer principle** alternating current in primary creates changing flux linking secondary, emf induced e = -N dΦ/dt, flux same through both, so V∝N. For N_p=550, N_s=1100, V_p=110 V rms, V_s= V_p×N_s/N_p=220 V, step-up factor 2, efficiency

Ref: NCERT > Physics Book > Alternating Currents > Transformer, AC Generator and LC Oscillations

A series LCR circuit has \( L = 1.5 \, \text{H} \), \( C = 35 \, \mu\text{F} \). What is the resonant frequency in Hz?

**Average power in LCR** P_avg = V_rms I_rms cos φ, cos φ = R/Z power factor, φ phase between V and I, tan φ = (X_L - X_C)/R. For R=80 Ω, X_L=100 Ω, X_C=40 Ω, X_L-X_C=60 Ω, Z=√(80²+60²)=100 Ω, cos φ=0.8, V_rms=240 V, I_rms=2.4 A, P=240×2.4×0.8=460.8 W, only R dissipates. Resonant angular frequency: ω₀ = (1/√(L C)) . L = 1.5 H , C = 35 × 10⁻⁶ F . ω₀ = (1/√(1.5 × 35 × 10⁻⁶)) ≈ 138.3 rad/s . f₀ = (ω₀/2π) = (138.3/6.28) ≈ 22 Hz . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms

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

A \( 200 \, \text{V} \) (rms) AC source is connected to a series LCR circuit with \( R = 20 \, \Omega \) at resonance. W

**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. At resonance, Z = R = 20 Ω . RMS current: I = (V/R) = (200/20) = 10 A . Power: P = I² R = 10² × 20 = 2000 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 2000

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

In an LCR circuit with \( R = 3 \, \Omega \), \( X_L = 8 \, \Omega \), \( X_C = 4 \, \Omega \), what is the power factor

**Average power in LCR** P_avg = V_rms I_rms cos φ, cos φ = R/Z power factor, φ phase between V and I, tan φ = (X_L - X_C)/R. For R=80 Ω, X_L=100 Ω, X_C=40 Ω, X_L-X_C=60 Ω, Z=√(80²+60²)=100 Ω, cos φ=0.8, V_rms=240 V, I_rms=2.4 A, P=240×2.4×0.8=460.8 W, only R dissipates. Impedance: Z = √(R² + (X_L - X_C)²) = √(3² + (8 - 4)²) = √(9 + 16) = 5 Ω . Power factor: cos Φ = (R/Z) = (3/5) = 0.6 . 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,

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

A series LCR circuit has \( L = 4 \, \text{H} \), \( C = 25 \, \mu\text{F} \). What is the resonant angular frequency?

**Condition for resistive behavior** X_L = X_C, so ωL=1/ωC, ω²=1/LC, this is resonance, impedance purely resistive Z=R, current maximum, power maximum P= V²/R, inductive and capacitive reactances equal magnitude opposite sign cancel, circuit appears resistive. ω₀ = (1/√(L C)) . L = 4 H , C = 25 × 10⁻⁶ F . ω₀ = (1/√(4 × 25 × 10⁻⁶)) = (1/√(10⁻⁴)) = 100 rad/s . 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 100 rad/s, consistent with phasor analysis and resonance condition X_L = X_C.

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

A series LCR circuit has \( R = 60 \, \Omega \), \( X_L = 50 \, \Omega \), \( X_C = 30 \, \Omega \). What is the impedan

**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. Z = √(R² + (X_L - X_C)²) . Z = √(60² + (50 - 30)²) = √(3600 + 400) = √(4000) ≈ 63.25 Ω . 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 62 Ω, consistent with phasor analysis and resonance condition X_L = X_C.

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

A series LCR circuit has \( R = 200 \, \Omega \), \( C = 15 \, \mu\text{F} \), and is connected to a \( 220 \, \text{V}

**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. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s , C = 15 × 10⁻⁶ F . X_C = (1/314 × 15 × 10⁻⁶) ≈ 212.3 Ω . No inductor, so X_L = 0 . Impedance: Z = √(R² + (X_C - X_L)²) = √(200² + 212.3²) ≈ 291.5 Ω . Applying X_L = ωL, X_C =

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

A series LCR circuit has \( L = 2.5 \, \text{H} \), \( C = 40 \, \mu\text{F} \). What is the resonant angular frequency?

**Current relative to voltage** in resistor in phase, φ=0°, power factor cos φ=1, maximum power, unlike inductor/capacitor where average power zero due to 90° phase shift, explaining why resistor heats while pure L/C does not. ω₀ = (1/√(L C)) . L = 2.5 H , C = 40 × 10⁻⁶ F . ω₀ = (1/√(2.5 × 40 × 10⁻⁶)) = (1/√(10⁻⁴)) = 100 rad/s . 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 100 rad/s, consistent with phasor analysis and resonance condition X_L = X_C.

Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power

A series LCR circuit with \( R = 50 \, \Omega \) is at resonance with a \( 250 \, \text{V} \) (rms) source. What is the

**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. At resonance, Z = R = 50 Ω . RMS current: I = (V/R) = (250/50) = 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 > Resonance in LCR Circuit and Q-Factor

A series LCR circuit has \( R = 80 \, \Omega \), \( X_L = 60 \, \Omega \), \( X_C = 40 \, \Omega \). What is the impedan

**LCR example** R=100 Ω, X_L=130 Ω, X_C=70 Ω, X_L-X_C=60 Ω, Z=√(100²+60²)=116.6 Ω, V_rms=300 V, I_rms=2.573 A, power P= I_rms² R =662 W? Actually P= V_rms I_rms cos φ, cos φ=R/Z=0.857, P=300×2.573×0.857=661.6 W, illustrating power factor. Z = √(R² + (X_L - X_C)²) . Z = √(80² + (60 - 40)²) = √(6400 + 400) = √(6800) ≈ 82.46 Ω . 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 82.46 Ω, consistent with phasor analysis and resonance condition X_L = X_C.

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

A series LCR circuit with \( R = 50 \, \Omega \), \( X_L = 70 \, \Omega \), \( X_C = 30 \, \Omega \) has a \( 200 \, \te

**LCR series impedance** Z = √(R² + (X_L - X_C)²), R resistance (Ω), X_L=ωL, X_C=1/ωC, phase angle φ = tan⁻¹((X_L-X_C)/R), current I_rms = V_rms/Z, voltage across R in phase with I, across L leads by 90°, across C lags by 90°, phasor diagram vector sum V = √(V_R² + (V_L - V_C)²). Z = √(R² + (X_L - X_C)²) = √(50² + (70 - 30)²) = √(2500 + 1600) = √(4100) ≈ 64 Ω . RMS current: I = (V/Z) = (200/64) ≈ 3.125 A . Power: P = I² R = (3.125)² × 50 ≈ 488.28 W . Applying X_L = ωL, X_C =

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

A series LCR circuit has \( L = 6 \, \text{H} \), \( C = 10 \, \mu\text{F} \). What is the resonant angular frequency?

**LCR example** R=100 Ω, X_L=130 Ω, X_C=70 Ω, X_L-X_C=60 Ω, Z=√(100²+60²)=116.6 Ω, V_rms=300 V, I_rms=2.573 A, power P= I_rms² R =662 W? Actually P= V_rms I_rms cos φ, cos φ=R/Z=0.857, P=300×2.573×0.857=661.6 W, illustrating power factor. ω₀ = (1/√(L C)) . L = 6 H , C = 10 × 10⁻⁶ F . ω₀ = (1/√(6 × 10 × 10⁻⁶)) = (1/√(6 × 10⁻⁵)) ≈ 129.1 rad/s . 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 129.1 rad/s, consistent with phasor analysis and resonance condition X_L = X_C.

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