#capacitive circuit

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

In an AC circuit with a series combination of resistor and capacitor, what determines whether the circuit is predominant

**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. In an RC series circuit, the circuit is predominantly capacitive if the capacitive reactance ( X_C = (1/ω C) ) is greater than the resistance ( R ). This makes the impedance Z = √(R² + X_C²) dominated by X_C , and the current leads the voltage significantly. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² +

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AC Through Inductor - Inductive Reactance

A \( 16 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) 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 = 16 × 10⁻⁶ F . X_C = (1/314 × 16 × 10⁻⁶) ≈ 199 Ω . RMS current: I = (V/X_C) = (230/199) ≈ 1.156 A . Peak current: i_m = √(2) I = 1.414 ×

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Power in AC Circuits - Power Factor and Wattless Current

What is the average power dissipated in a purely capacitive circuit over one complete cycle?

**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. Power: p_C = i v = i_m v_m cos (ω t) sin (ω t) = (i_m v_m/2) sin (2ω t) . Average over a cycle: P_C = (i_m v_m/2) langle sin (2ω t) rangle = 0 , since langle sin (2ω t) rangle = 0 . Applying X_L = ωL, X_C =

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AC Through Capacitor - Capacitive Reactance

In a purely capacitive AC circuit, the current leads the voltage by what phase angle?

**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. For a capacitor, i = i_m sin (ω t + (π/2)) , v = v_m sin ω t . Phase difference: Φ = (π/2) , so current leads voltage by 90° . 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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