#capacitance

Power in AC Circuits - Power Factor and Wattless Current

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

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Resonance in LCR Circuit and Q-Factor

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.

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Resonance in LCR Circuit and Q-Factor

A series LCR circuit with \( R = 100 \, \Omega \), \( L = 1 \, \text{H} \), \( C = 1 \, \mu\text{F} \) is at resonance.

**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. At resonance, X_L = X_C , so Z = R . Given: R = 100 Ω . Impedance Z = 100 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 100 Ω, consistent with phasor analysis and resonance condition X_L = X_C.

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AC Through Resistor - Phasor and Power

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.

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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.

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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

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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( L = 3 \, \text{H} \), \( C = 12 \, \mu\text{F} \). 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 = 3 H , C = 12 × 10⁻⁶ F . ω₀ = (1/√(3 × 12 × 10⁻⁶)) = (1/√(36 × 10⁻⁶)) = (10³/6) ≈ 166.67 rad/s . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P =

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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( L = 5 \, \text{H} \), \( C = 20 \, \mu\text{F} \). 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 = 5 H , C = 20 × 10⁻⁶ F . ω₀ = (1/√(5 × 20 × 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.

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AC Through Inductor - Inductive Reactance

A \( 20 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC supply. What is th

**Inductive reactance** X_L = ω L =2π f L (Ω), L inductance (H), f frequency (Hz), ω=2πf angular frequency (rad/s), opposes AC, impedance Z = X_L for pure L, current lags voltage by 90°, I_rms = V_rms/X_L, I_peak = V_peak/X_L. For 85 mH, 50 Hz, X_L=2π×50×0.085=26.7 Ω, V_rms=230 V, I_rms=8.61 A, I_peak=12.18 A. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 20 × 10⁻⁶ F . X_C = (1/314 × 20 × 10⁻⁶) ≈ 159.2 Ω . RMS current: I = (V/X_C) = (220/159.2) ≈ 1.38 A . Applying X_L = ωL, X_C = 1/ωC, Z =

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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( L = 2 \, \text{H} \), \( C = 50 \, \mu\text{F} \). What is the resonant angular frequency?

**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. Resonant frequency: ω₀ = (1/√(L C)) . L = 2 H , C = 50 × 10⁻⁶ F . ω₀ = (1/√(2 × 50 × 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.

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AC Through Capacitor - Capacitive Reactance

In an AC circuit with only a capacitor, what happens to the impedance if the capacitance is decreased?

**Capacitor average power** zero over complete cycle because P=V I =½ V_peak I_peak sin2ωt average zero, energy stored in field, not dissipated, unlike resistor. For 15 μF, 60 Hz, X_C=176.8 Ω, V_rms=110 V, I_rms=0.622 A, illustrating lower C higher X_C. In a purely capacitive circuit, impedance is X_C = (1/ω C) . Decreasing capacitance ( C ) increases X_C because they are inversely proportional, thus increasing the impedance. 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 increases, consistent with phasor analysis and resonance condition X_L = X_C.

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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( R = 40 \, \Omega \), \( L = 5 \, \text{H} \), \( C = 80 \, \mu\text{F} \). What is the reson

**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. ω₀ = (1/√(L C)) . L = 5 H , C = 80 × 10⁻⁶ F . ω₀ = (1/√(5 × 80 × 10⁻⁶)) = (1/√(4 × 10⁻⁴)) = 50 rad/s . f₀ = (50/2 × 3.14) ≈ 7.96 Hz . 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 8 Hz, consistent with

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