AC Fundamentals - RMS, Average and Peak Values

AC Fundamentals - RMS, Average and Peak Values

A \( 25 \, \Omega \) resistor is connected to a \( 125 \, \text{V} \) (rms) AC source. What is the rms current?

**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 current: I = (V/R) . Given: V = 125 V , R = 25 Ω . I = (125/25) = 5 A . 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 5 A, consistent with phasor analysis and resonance condition X_L = X_C.

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AC Fundamentals - RMS, Average and Peak Values

A series LCR circuit with \( R = 70 \, \Omega \), \( X_L = 90 \, \Omega \), \( X_C = 40 \, \Omega \) has a \( 210 \, \te

**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. Z = √(R² + (X_L - X_C)²) = √(70² + (90 - 40)²) = √(4900 + 2500) = √(7400) ≈ 86 Ω . RMS current: I = (V/Z) = (210/86) ≈ 2.442 A . Power: P = I² R = (2.442)² × 70 ≈ 417.6 W . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P =

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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

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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 \( 180 \, \text{V} \) (rms) source supplies a \( 90 \, \Omega \) resistor. What is the peak current?

**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. RMS current: I = (V/R) = (180/90) = 2 A . Peak current: i_m = √(2) I = 1.414 × 2 ≈ 2.828 A . 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 2.828 A, consistent with phasor analysis and resonance condition X_L = X_C.

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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 =

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AC Fundamentals - RMS, Average and Peak Values

A \( 50 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) 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 = 50 × 10⁻⁶ F . X_C = (1/314 × 50 × 10⁻⁶) ≈ 63.7 Ω . RMS current: I = (V/X_C) = (220/63.7) ≈ 3.45 A . Peak current: i_m = √(2) I = 1.414 × 3.45 ≈ 4.88 A . Applying X_L =

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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

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AC Fundamentals - RMS, Average and Peak Values

A \( 35 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is the r

**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. X_L = ω L , ω = 2π × 60 = 376.8 rad/s . L = 35 × 10⁻³ H . X_L = 376.8 × 0.035 = 13.19 Ω . RMS current: I = (V/X_L) = (110/13.19) ≈ 8.34 A . 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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AC Fundamentals - RMS, Average and Peak Values

A \( 100 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the

**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. X_L = ω L , ω = 2π f . f = 50 Hz , L = 100 × 10⁻³ H . ω = 2 × 3.14 × 50 = 314 rad/s . X_L = 314 × 0.1 = 31.4 Ω . 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 Fundamentals - RMS, Average and Peak Values

A series LCR circuit with \( R = 90 \, \Omega \), \( X_L = 120 \, \Omega \), \( X_C = 60 \, \Omega \) has a \( 270 \, \t

**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. Z = √(R² + (X_L - X_C)²) = √(90² + (120 - 60)²) = √(8100 + 3600) = √(11700) ≈ 108.17 Ω . RMS current: I = (V/Z) = (270/108.17) ≈ 2.496 A . Power: P = I² R = (2.496)² × 90 ≈ 560.6 W . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P =

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AC Fundamentals - RMS, Average and Peak Values

In a purely inductive AC circuit, the current lags the voltage by what phase angle?

**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. For a pure inductor, i = i_m sin (ω t - (π/2)) , while v = v_m sin ω t . Phase difference: Φ = -(π/2) , meaning current lags voltage by (π/2) or 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 =

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