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#average power

17 public questions tagged with this topic.

Power in AC Circuits - Power Factor and Wattless Current

A \( 180 \, \text{V} \) (rms) AC source supplies a \( 90 \, \Omega \) resistor. What is the average power consumed?

**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. RMS current: I = (V/R) = (180/90) = 2 A . Average power: P = I² R = 2² × 90 = 360 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

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Power in AC Circuits - Power Factor and Wattless Current

A \( 268.7 \, \text{V} \) (peak) AC source is connected to a \( 95 \, \Omega \) resistor. What is the average power cons

**Wattless current** occurs in pure inductor or capacitor, I_rms non-zero but average power zero because φ=±90°, cos φ=0, energy oscillates between source and field, no dissipation, used in choke coil to limit current without heating, unlike resistor where power dissipated. RMS voltage: V = (v_m/√(2)) = (268.7/1.414) ≈ 190 V . RMS current: I = (V/R) = (190/95) = 2 A . Average power: P = I² R = 2² × 95 = 380 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 380 W,

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Power in AC Circuits - Power Factor and Wattless Current

A \( 165 \, \text{V} \) (rms) AC source supplies a \( 55 \, \Omega \) resistor. What is the average power consumed?

**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. RMS current: I = (V/R) = (165/55) = 3 A . Average power: P = I² R = 3² × 55 = 495 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

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AC Through Resistor - Phasor and Power

A \( 169.7 \, \text{V} \) (peak) AC source is connected to a \( 60 \, \Omega \) resistor. What is the average power cons

**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. RMS voltage: V = (v_m/√(2)) = (169.7/1.414) ≈ 120 V . RMS current: I = (V/R) = (120/60) = 2 A . Average power: P = I² R = 2² × 60 = 240 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 240 W, consistent with phasor analysis and resonance condition X_L = X_C.

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Power in AC Circuits - Power Factor and Wattless Current

In an AC circuit with a pure inductor, why is the average power dissipated zero over a 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. In a pure inductor, the current lags the voltage by 90°. The instantaneous power oscillates between positive (energy stored) and negative (energy returned), averaging to zero over a cycle because the inductor does not dissipate energy as heat but stores and releases it. Applying X_L = ωL, X_C = 1/ωC, Z =

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Power in AC Circuits - Power Factor and Wattless Current

A \( 175 \, \text{V} \) (rms) AC source supplies a \( 35 \, \Omega \) resistor. What is the average power consumed?

**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. RMS current: I = (V/R) = (175/35) = 5 A . Average power: P = I² R = 5² × 35 = 875 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 875 W, consistent

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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 \( 130 \, \text{V} \) (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 Resistor - Phasor and Power

A \( 141.4 \, \text{V} \) (peak) AC source is connected to a \( 50 \, \Omega \) resistor. What is the average power cons

**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. RMS voltage: V = (v_m/√(2)) = (141.4/1.414) = 100 V . RMS current: I = (V/R) = (100/50) = 2 A . Average power: P = I² R = 2² × 50 = 200 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 200 W, consistent with phasor analysis and resonance condition X_L = X_C.

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AC Fundamentals - RMS, Average and Peak Values

A \( 212.1 \, \text{V} \) (peak) AC source is connected to a \( 75 \, \Omega \) resistor. What is the average power cons

**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. RMS voltage: V = (v_m/√(2)) = (212.1/1.414) = 150 V . RMS current: I = (V/R) = (150/75) = 2 A . Average power: P = I² R = 2² × 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 W, consistent

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

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AC Fundamentals - RMS, Average and Peak Values

A \( 198.1 \, \text{V} \) (peak) AC source is connected to a \( 70 \, \Omega \) resistor. What is the average power cons

**Peak current** I_peak = V_peak/R for resistor, I_rms = V_rms/R, V_peak = √2 V_rms, for 200 V rms, V_peak=282.8 V, I_peak=282.8/80=3.535 A, rms I=200/80=2.5 A, average over complete cycle zero because positive and negative halves cancel. RMS voltage: V = (v_m/√(2)) = (198.1/1.414) ≈ 140 V . RMS current: I = (V/R) = (140/70) = 2 A . Average power: P = I² R = 2² × 70 = 280 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 280 W, consistent with phasor analysis

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

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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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