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#peak voltage

7 public questions tagged with this topic.

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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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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AC Through Resistor - Phasor and Power

A \( 200 \, \text{V} \) (rms) AC source supplies a \( 100 \, \Omega \) resistor. What is the peak voltage?

**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. Peak voltage: v_m = √(2) V . V = 200 V . v_m = 1.414 × 200 = 282.8 V ≈ 283 V . 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 283 V, 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 \( 198.1 \, \text{V} \) (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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AC Fundamentals - RMS, Average and Peak Values

An AC source provides a peak voltage of \( 424.2 \, \text{V} \) to a \( 200 \, \Omega \) resistor. What is the average p

**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)) = (424.2/1.414) = 300 V . RMS current: I = (V/R) = (300/200) = 1.5 A . Average power: P = I² R = (1.5)² × 200 = 2.25 × 200 = 450 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

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 \( 254.6 \, \text{V} \) (peak) AC source is connected to a \( 90 \, \Omega \) 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)) = (254.6/1.414) ≈ 180 V . 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 360 W, consistent

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

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