AC Through Capacitor - Capacitive Reactance

AC Through Capacitor - Capacitive Reactance

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

**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, the capacitive reactance ( X_C = (1/ω C) ) decreases as the frequency ( f , where ω = 2π f ) increases. Since current is inversely proportional to reactance ( I = (V/X_C) ), the current increases. 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,

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AC Through Capacitor - Capacitive Reactance

A \( 16 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the c

**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. X_C = (1/ω C) , ω = 2π × 60 = 376.8 rad/s . C = 16 × 10⁻⁶ F . X_C = (1/376.8 × 16 × 10⁻⁶) ≈ 165.8 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

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AC Through Capacitor - Capacitive Reactance

A \( 28 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the r

**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. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 28 × 10⁻⁶ F . X_C = (1/314 × 28 × 10⁻⁶) ≈ 113.6 Ω . RMS current: I = (V/X_C) = (220/113.6) ≈ 1.936 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,

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AC Through Capacitor - Capacitive Reactance

A \( 50 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the peak

**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. X_L = ω L , ω = 2π × 50 = 314 rad/s . L = 50 × 10⁻³ H . X_L = 314 × 0.05 = 15.7 Ω . RMS current: I = (V/X_L) = (220/15.7) ≈ 14.01 A . Peak current: i_m = √(2) I = 1.414 × 14.01 ≈ 19.81 A . Applying X_L = ωL,

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AC Through Capacitor - Capacitive Reactance

A \( 150 \, \text{V} \) (rms) AC source supplies a \( 75 \, \Omega \) resistor. What is the average power consumed?

**Capacitive reactance** X_C =1/(ω C)=1/(2π f C) (Ω), C capacitance (F), current leads voltage by 90°, I_rms = V_rms/X_C = V_rms ω C, I_peak = V_peak ω C, impedance Z = X_C for pure C. For 45 μF, 60 Hz, X_C=1/(2π×60×45×10⁻⁶)=58.9 Ω, V_rms=110 V, I_rms=1.867 A, I_peak=2.64 A. RMS current: I = (V/R) = (150/75) = 2 A . Average power: P = I² R = 2² × 75 = 4 × 75 = 300 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 300

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AC Through Capacitor - Capacitive Reactance

A \( 30 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the rms

**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. X_L = ω L , ω = 2π × 50 = 314 rad/s . L = 30 × 10⁻³ H . X_L = 314 × 0.03 = 9.42 Ω . RMS current: I = (V/X_L) = (220/9.42) ≈ 23.35 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 =

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AC Through Capacitor - Capacitive Reactance

A \( 40 \, \Omega \) resistor is connected to a \( 120 \, \text{V} \) (rms) AC source. What is the rms current?

**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) . Given: V = 120 V , R = 40 Ω . I = (120/40) = 3 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 3 A, consistent with phasor analysis and resonance condition X_L = X_C.

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AC Through Capacitor - Capacitive Reactance

A \( 6 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the ca

**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. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 6 × 10⁻⁶ F . X_C = (1/314 × 6 × 10⁻⁶) ≈ 530.5 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

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AC Through Capacitor - Capacitive Reactance

A \( 25 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is th

**Capacitive reactance** X_C =1/(ω C)=1/(2π f C) (Ω), C capacitance (F), current leads voltage by 90°, I_rms = V_rms/X_C = V_rms ω C, I_peak = V_peak ω C, impedance Z = X_C for pure C. For 45 μF, 60 Hz, X_C=1/(2π×60×45×10⁻⁶)=58.9 Ω, V_rms=110 V, I_rms=1.867 A, I_peak=2.64 A. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 25 × 10⁻⁶ F . X_C = (1/314 × 25 × 10⁻⁶) ≈ 127.4 Ω . RMS current: I = (V/X_C) = (230/127.4) ≈ 1.805 A . Peak current: i_m = √(2) I = 1.414 × 1.805 ≈ 2.55 A .

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AC Through Capacitor - Capacitive Reactance

A \( 11 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the c

**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. X_C = (1/ω C) , ω = 2π × 60 = 376.8 rad/s . C = 11 × 10⁻⁶ F . X_C = (1/376.8 × 11 × 10⁻⁶) ≈ 241.2 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

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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.

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AC Through Capacitor - Capacitive Reactance

A series LCR circuit with \( R = 60 \, \Omega \), \( X_L = 80 \, \Omega \), \( X_C = 20 \, \Omega \) has a \( 180 \, \te

**Capacitive reactance** X_C =1/(ω C)=1/(2π f C) (Ω), C capacitance (F), current leads voltage by 90°, I_rms = V_rms/X_C = V_rms ω C, I_peak = V_peak ω C, impedance Z = X_C for pure C. For 45 μF, 60 Hz, X_C=1/(2π×60×45×10⁻⁶)=58.9 Ω, V_rms=110 V, I_rms=1.867 A, I_peak=2.64 A. Z = √(R² + (X_L - X_C)²) = √(60² + (80 - 20)²) = √(3600 + 3600) = √(7200) ≈ 84.85 Ω . RMS current: I = (V/Z) = (180/84.85) ≈ 2.12 A . Power: P = I² R = (2.12)² × 60 ≈ 269.66 W . Applying X_L = ωL, X_C = 1/ωC, Z = √(R²

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