#series lcr circuit

Power in AC Circuits - Power Factor and Wattless Current

A series LCR circuit has \( R = 10 \, \Omega \), \( X_L = 15 \, \Omega \), \( X_C = 5 \, \Omega \). What is the power fa

**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. Z = √(R² + (X_L - X_C)²) = √(10² + (15 - 5)²) = √(100 + 100) = 14.14 Ω . Power factor: cos Φ = (R/Z) = (10/14.14) ≈ 0.707 . 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

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AC Fundamentals - RMS, Average and Peak Values

A series LCR circuit with \( R = 90 \, \Omega \), \( X_L = 120 \, \Omega \), \( X_C = 60 \, \Omega \) has a \( 270 \, \t

**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. Z = √(R² + (X_L - X_C)²) = √(90² + (120 - 60)²) = √(8100 + 3600) = √(11700) ≈ 108.17 Ω . RMS current: I = (V/Z) = (270/108.17) ≈ 2.496 A . Power: P = I² R = (2.496)² × 90 ≈ 560.6 W . 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 > AC Fundamentals - RMS, Average and Peak Values

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Power in AC Circuits - Power Factor and Wattless Current

A series LCR circuit has \( R = 40 \, \Omega \), \( X_L = 65 \, \Omega \), \( X_C = 25 \, \Omega \). 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) = √(40² + (65 - 25)²) = √(1600 + 1600) = √(3200) ≈ 56.57 Ω . Power factor: cos Φ = (R/Z) = (40/56.57) ≈ 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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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( L = 8 \, \text{H} \), \( C = 5 \, \mu\text{F} \). What is the resonant angular frequency?

**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). ω₀ = (1/√(L C)) . L = 8 H , C = 5 × 10⁻⁶ F . ω₀ = (1/√(8 × 5 × 10⁻⁶)) = (1/√(40 × 10⁻⁶)) ≈ 158.1 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

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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( L = 4 \, \text{H} \), \( C = 10 \, \mu\text{F} \). What is the resonant angular frequency?

**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). ω₀ = (1/√(L C)) . L = 4 H , C = 10 × 10⁻⁶ F . ω₀ = (1/√(4 × 10 × 10⁻⁶)) = (1/√(4 × 10⁻⁵)) ≈ 158.1 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

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

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