AC Through Resistor - Phasor and Power

AC Through Resistor - Phasor and Power

A \( 60 \, \Omega \) resistor and \( 15 \, \mu\text{F} \) capacitor are in series with a \( 230 \, \text{V} \), \( 50 \,

**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. X_C = (1/ω C) = (1/314 × 15 × 10⁻⁶) ≈ 212.3 Ω . Z = √(R² + X_C²) = √(60² + 212.3²) = √(3600 + 45071.29) ≈ 220.8 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 220.8 Ω, consistent with phasor analysis and resonance condition X_L = X_C.

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AC Through Resistor - Phasor and Power

A \( 13 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the c

**Resistor in AC** behaves as DC, no reactance, impedance Z=R, current follows voltage exactly, average power over cycle V_rms I_rms, for 200 V rms, 80 Ω, P=500 W? Actually 200²/80=500 W, peak current √2×2.5=3.535 A, average power ½ V_peak I_peak. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 13 × 10⁻⁶ F . X_C = (1/314 × 13 × 10⁻⁶) ≈ 245.1 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 245.1 Ω, consistent with phasor analysis

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AC Through Resistor - Phasor and Power

A series LCR circuit has \( L = 2.5 \, \text{H} \), \( C = 40 \, \mu\text{F} \). What is the resonant angular frequency?

**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. ω₀ = (1/√(L C)) . L = 2.5 H , C = 40 × 10⁻⁶ F . ω₀ = (1/√(2.5 × 40 × 10⁻⁶)) = (1/√(10⁻⁴)) = 100 rad/s . 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 100 rad/s, consistent with phasor analysis and resonance condition X_L = X_C.

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AC Through Resistor - Phasor and Power

A \( 50 \, \Omega \) resistor and \( 20 \, \mu\text{F} \) capacitor are in series with a \( 110 \, \text{V} \), \( 50 \,

**AC through resistor** voltage and current in phase, φ=0°, I = V/R instantaneously, I(t)=I_peak sin ωt, V(t)=V_peak sin ωt, phasor diagram V and I same direction, power instantaneous P = V I = V_peak I_peak sin² ωt, average P_avg = V_rms I_rms = V_rms²/R, always positive, energy dissipated as heat. X_C = (1/ω C) = (1/314 × 20 × 10⁻⁶) ≈ 159.2 Ω . Z = √(R² + X_C²) = √(50² + 159.2²) ≈ 166.6 Ω . RMS current: I = (V/Z) = (110/166.6) ≈ 0.66 A . Voltage across resistor: V_R = I R = 0.66 × 50 ≈ 33 V . Applying X_L

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AC Through Resistor - Phasor and Power

A \( 240 \, \text{V} \) (rms) source supplies a \( 120 \, \Omega \) resistor. What is the peak current?

**Resistor in AC** behaves as DC, no reactance, impedance Z=R, current follows voltage exactly, average power over cycle V_rms I_rms, for 200 V rms, 80 Ω, P=500 W? Actually 200²/80=500 W, peak current √2×2.5=3.535 A, average power ½ V_peak I_peak. RMS current: I = (V/R) = (240/120) = 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 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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AC Through Resistor - Phasor and Power

A \( 15 \, \Omega \) resistor is connected to a \( 75 \, \text{V} \) (rms) AC source. What is the rms current?

**Resistor in AC** behaves as DC, no reactance, impedance Z=R, current follows voltage exactly, average power over cycle V_rms I_rms, for 200 V rms, 80 Ω, P=500 W? Actually 200²/80=500 W, peak current √2×2.5=3.535 A, average power ½ V_peak I_peak. RMS current: I = (V/R) . Given: V = 75 V , R = 15 Ω . I = (75/15) = 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 Through Resistor - Phasor and Power

Why does an ideal transformer maintain constant power across its primary and secondary coils?

**AC through resistor** voltage and current in phase, φ=0°, I = V/R instantaneously, I(t)=I_peak sin ωt, V(t)=V_peak sin ωt, phasor diagram V and I same direction, power instantaneous P = V I = V_peak I_peak sin² ωt, average P_avg = V_rms I_rms = V_rms²/R, always positive, energy dissipated as heat. In an ideal transformer, there are no losses (e.g., resistance, flux leakage). Energy conservation dictates that input power ( V_p I_p ) equals output power ( V_s I_s ), so the power remains constant despite changes in voltage and current. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_

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AC Through Resistor - Phasor and Power

A \( 95 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the peak

**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. X_L = ω L , ω = 2π × 50 = 314 rad/s . L = 95 × 10⁻³ H . X_L = 314 × 0.095 = 29.83 Ω . RMS current: I = (V/X_L) = (220/29.83) ≈ 7.375 A . Peak current: i_m = √(2) I = 1.414 × 7.375 ≈ 10.43 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms

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AC Through Resistor - Phasor and Power

A \( 190 \, \text{V} \) (rms) source supplies a \( 95 \, \Omega \) resistor. What is the peak current?

**AC through resistor** voltage and current in phase, φ=0°, I = V/R instantaneously, I(t)=I_peak sin ωt, V(t)=V_peak sin ωt, phasor diagram V and I same direction, power instantaneous P = V I = V_peak I_peak sin² ωt, average P_avg = V_rms I_rms = V_rms²/R, always positive, energy dissipated as heat. RMS current: I = (V/R) = (190/95) = 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

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AC Through Resistor - Phasor and Power

A \( 50 \, \Omega \) resistor and \( 20 \, \mu\text{F} \) capacitor are in series with a \( 230 \, \text{V} \), \( 50 \,

**AC through resistor** voltage and current in phase, φ=0°, I = V/R instantaneously, I(t)=I_peak sin ωt, V(t)=V_peak sin ωt, phasor diagram V and I same direction, power instantaneous P = V I = V_peak I_peak sin² ωt, average P_avg = V_rms I_rms = V_rms²/R, always positive, energy dissipated as heat. X_C = (1/ω C) = (1/314 × 20 × 10⁻⁶) ≈ 159.2 Ω . Z = √(R² + X_C²) = √(50² + 159.2²) = √(2500 + 25344.64) ≈ 166.6 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 =

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AC Through Resistor - Phasor and Power

What is the effect on an ideal transformer’s secondary voltage if the number of turns in the secondary coil is halved?

**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 an ideal transformer, (V_s/V_p) = (N_s/N_p) . If the number of secondary turns ( N_s ) is halved, the secondary voltage ( V_s ) becomes half its original value, assuming primary voltage ( V_p ) remains constant. 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 It halves, consistent wit

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