#power calculation

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,

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

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

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

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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.

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

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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

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

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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.

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

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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