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#reactive power

11 public questions tagged with this topic.

Power in AC Circuits - Power Factor and Wattless Current

A \( 60 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC supply. What is th

**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. Capacitive reactance: X_C = (1/ω C) , where ω = 2π f . f = 60 Hz , C = 60 × 10⁻⁶ F . ω = 2 × 3.14 × 60 = 376.8 rad/s . X_C = (1/376.8 × 60 × 10⁻⁶) = 44.24 Ω . RMS current: I = (V/X_C)

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Power in AC Circuits - Power Factor and Wattless Current

What is the significance of the phase difference being 90° in a purely reactive AC circuit?

**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. A 90° phase difference (in purely inductive or capacitive circuits) means the power factor ( cos 90° = 0 ) is zero, indicating no average power is dissipated. Energy oscillates between the source and the reactive element without being converted to heat. Applying X_L = ωL, X_C = 1/ωC, Z = √(R²

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AC Through Capacitor - Capacitive Reactance

In an AC circuit with only a capacitor, what is the nature of the current when the voltage is at its peak?

**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. In a purely capacitive circuit, the current leads the voltage by 90°. When the voltage is at its peak (maximum positive or negative), the current is zero because it reaches its peak 90° earlier and crosses zero at the voltage peak. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms =

Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance

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AC Fundamentals - RMS, Average and Peak Values

Why does a purely capacitive AC circuit not dissipate power despite having current flow?

**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. In a purely capacitive circuit, the current leads the voltage by 90°. The instantaneous power oscillates, but the average power over a cycle is zero because the energy is stored during one half-cycle and returned during the other, with no net dissipation. 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 \( 20 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) 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 = 20 × 10⁻⁶ F . X_C = (1/314 × 20 × 10⁻⁶) ≈ 159.2 Ω . RMS current: I = (V/X_C) = (230/159.2) ≈ 1.445 A . Peak current: i_m = √(2) I = 1.414 × 1.445 ≈ 2.04 A . Applying X_L =

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

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Resonance in LCR Circuit and Q-Factor

Why does the voltage across an inductor in a series LCR circuit potentially exceed the source voltage 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, current is maximized ( I = (V/R) ), and the voltage across the inductor ( V_L = I X_L ) can exceed the source voltage if X_L > R . This is possible because V_L and V_C are out of phase with the source voltage, and their vector sum with V_R equals the source voltage. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L -

Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor

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Power in AC Circuits - Power Factor and Wattless Current

Why does an AC circuit with only an inductor exhibit a "wattless" current?

**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. In a purely inductive AC circuit, the current lags the voltage by 90°. The power factor ( cos 90° = 0 ) is zero, meaning no average power is dissipated; the current is "wattless" because it only stores and releases energy without consuming it. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² +

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Power in AC Circuits - Power Factor and Wattless Current

What is the role of a capacitor in improving the power factor of an AC circuit with an inductive load?

**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. An inductive load causes the current to lag the voltage, reducing the power factor. Adding a capacitor in parallel introduces a leading current that counteracts the lagging current, bringing the net phase difference closer to zero, thus improving the power factor. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² +

Ref: NCERT > Physics Book > Alternating Currents > Power in AC Circuits - Power Factor and Wattless Current

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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

What happens to the average power dissipated in an AC circuit with only a capacitor over one complete cycle?

**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. In a purely capacitive AC circuit, the current leads the voltage by 90°. The instantaneous power alternates between positive and negative, and over a complete cycle, these values cancel out, resulting in zero average power dissipation. The capacitor stores and releases energy without dissipating it as heat. Applying X_L = ωL, X_C = 1/ωC, Z = √(R²

Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance

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AC Through Resistor - Phasor and Power

In an AC circuit with only an inductor, what is the instantaneous power when the current is at its maximum?

**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. In a purely inductive circuit, current lags voltage by 90°. When the current is at its maximum, the voltage is zero (since voltage leads by 90°), making the instantaneous power ( P = V I ) zero at that instant. 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 Zero, consistent with

Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power

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