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#capacitive reactance

10 public questions tagged with this topic.

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 Fundamentals - RMS, Average and Peak Values

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

**RMS value** I_rms = I_peak/√2, V_rms = V_peak/√2 for sinusoidal AC, significance rms gives equivalent DC value producing same heating power P = I_rms² R, average power over cycle, instruments measure rms, average over full cycle zero, half-cycle average 2 I_peak/π, peak = √2 rms. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 50 × 10⁻⁶ F . X_C = (1/314 × 50 × 10⁻⁶) ≈ 63.7 Ω . RMS current: I = (V/X_C) = (220/63.7) ≈ 3.45 A . Peak current: i_m = √(2) I = 1.414 × 3.45 ≈ 4.88 A . Applying X_L =

Ref: NCERT > Physics Book > Alternating Currents > AC Fundamentals - RMS, Average and Peak Values

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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.

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AC Through Capacitor - Capacitive Reactance

A \( 40 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the c

**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 = 40 × 10⁻⁶ F . X_C = (1/314 × 40 × 10⁻⁶) ≈ 79.6 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 79.6 Ω, consistent with phasor analysis

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AC Through Capacitor - Capacitive Reactance

A \( 15 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the c

**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 = 15 × 10⁻⁶ F . X_C = (1/314 × 15 × 10⁻⁶) ≈ 212 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 212 Ω, consistent with phasor analysis

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AC Through Capacitor - Capacitive Reactance

A \( 8 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) 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 × 60 = 376.8 rad/s . C = 8 × 10⁻⁶ F . X_C = (1/376.8 × 8 × 10⁻⁶) ≈ 331.6 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 Inductor - Inductive Reactance

A \( 30 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. 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 = 30 × 10⁻⁶ F . X_C = (1/314 × 30 × 10⁻⁶) ≈ 106.1 Ω . RMS current: I = (V/X_C) = (220/106.1) ≈ 2.074 A . Peak current: i_m = √(2) I = 1.414 ×

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AC Through Capacitor - Capacitive Reactance

A \( 9 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) 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 = 9 × 10⁻⁶ F . X_C = (1/314 × 9 × 10⁻⁶) ≈ 353.7 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 \( 12 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the c

**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 = 12 × 10⁻⁶ F . X_C = (1/314 × 12 × 10⁻⁶) ≈ 265.3 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 265.3 Ω, consistent with phasor analysis

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Resonance in LCR Circuit and Q-Factor

In a series LCR circuit at resonance, what is the relationship between the inductive and capacitive reactances?

**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 in a series LCR circuit, the inductive reactance ( X_L ) equals the capacitive reactance ( X_C ). This balance cancels out the reactive components, making the impedance purely resistive and maximizing the current. 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,

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