#series LCR

Power in AC Circuits - Power Factor and Wattless Current

In an AC circuit with a series LCR combination, why does the power dissipated depend only on the resistive component?

**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. Power dissipation in an AC circuit ( P = I² R cos Φ ) occurs only through resistance, as inductors and capacitors store and release energy without converting it to heat. The reactive components affect the current and phase, but only R dissipates power. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² +

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

View Details

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

View Details

LCR Series Circuit - Impedance and Phasor Diagram

What is the primary factor determining the resonant frequency in a series LCR circuit?

**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). The resonant frequency in a series LCR circuit is determined by the inductance ( L ) and capacitance ( C ), given by f₀ = (1/2π √(L C)) . Resistance affects damping but not the resonant frequency itself. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms

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

View Details

LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( R = 35 \, \Omega \), \( X_L = 55 \, \Omega \), \( X_C = 75 \, \Omega \). What is the phase a

**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. tan Φ = (X_C - X_L/R) = (75 - 55/35) = (20/35) ≈ 0.571 . Φ = tan⁻¹(0.571) ≈ 29.74° . Since X_C > X_L , current leads 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 gives 29.74°, consistent with phasor analysis and resonance condition X_L = X_C.

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

View Details

LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit with \( R = 50 \, \Omega \), \( X_L = 30 \, \Omega \), \( X_C = 20 \, \Omega \) is connected to an

**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. Impedance: Z = √(R² + (X_L - X_C)²) . Z = √(50² + (30 - 20)²) = √(2500 + 100) = √(2600) ≈ 51 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 50 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

View Details

Resonance in LCR Circuit and Q-Factor

In a series LCR circuit at resonance, the power dissipated is \( 4800 \, \text{W} \) with \( R = 3 \, \Omega \). What is

**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 , and power P = I² R . 4800 = I² × 3 . I² = (4800/3) = 1600 . I = √(1600) = 40 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

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

View Details

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

View Details