Power in AC Circuits - Power Factor and Wattless Current

Power in AC Circuits - Power Factor and Wattless Current

A \( 45 \, \text{mH} \) inductor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the peak

**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. X_L = ω L , ω = 2π × 50 = 314 rad/s . L = 45 × 10⁻³ H . X_L = 314 × 0.045 = 14.13 Ω . RMS current: I = (V/X_L) = (230/14.13) ≈ 16.28 A . Peak current: i_m = √(2) I = 1.414 × 16.28 ≈ 23.02 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R²

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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 \( 23 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the r

**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. X_C = (1/ω C) , ω = 2π × 60 = 376.8 rad/s . C = 23 × 10⁻⁶ F . X_C = (1/376.8 × 23 × 10⁻⁶) ≈ 115.4 Ω . RMS current: I = (V/X_C) = (110/115.4) ≈ 0.953 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L

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Power in AC Circuits - Power Factor and Wattless Current

A \( 70 \, \Omega \) resistor and \( 14 \, \mu\text{F} \) capacitor are in series with a \( 210 \, \text{V} \), \( 50 \,

**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. X_C = (1/ω C) = (1/314 × 14 × 10⁻⁶) ≈ 227.5 Ω . Z = √(R² + X_C²) = √(70² + 227.5²) = √(4900 + 51756.25) ≈ 238.2 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 238.2 Ω, consistent with phasor analysis and resonance condition X_L = X_C.

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Power in AC Circuits - Power Factor and Wattless Current

A \( 60 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC supply. What is th

**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. Capacitive reactance: X_C = (1/ω C) , where ω = 2π f . f = 60 Hz , C = 60 × 10⁻⁶ F . ω = 2 × 3.14 × 60 = 376.8 rad/s . X_C = (1/376.8 × 60 × 10⁻⁶) = 44.24 Ω . RMS current: I = (V/X_C)

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Power in AC Circuits - Power Factor and Wattless Current

A series LCR circuit has \( L = 1.5 \, \text{H} \), \( C = 35 \, \mu\text{F} \). What is the resonant frequency in Hz?

**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. Resonant angular frequency: ω₀ = (1/√(L C)) . L = 1.5 H , C = 35 × 10⁻⁶ F . ω₀ = (1/√(1.5 × 35 × 10⁻⁶)) ≈ 138.3 rad/s . f₀ = (ω₀/2π) = (138.3/6.28) ≈ 22 Hz . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms

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Power in AC Circuits - Power Factor and Wattless Current

In an AC circuit containing only a resistor, what happens to the power dissipated if the frequency of the source is doub

**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. In a purely resistive AC circuit, power dissipated is P = I² R , where I = (V/R) , and R is constant. Since resistance does not depend on frequency, and assuming the rms voltage remains constant, the power dissipated remains unchanged when frequency doubles. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms

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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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Power in AC Circuits - Power Factor and Wattless Current

In an LCR circuit with \( R = 3 \, \Omega \), \( X_L = 8 \, \Omega \), \( X_C = 4 \, \Omega \), what is the power factor

**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. Impedance: Z = √(R² + (X_L - X_C)²) = √(3² + (8 - 4)²) = √(9 + 16) = 5 Ω . Power factor: cos Φ = (R/Z) = (3/5) = 0.6 . 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,

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Power in AC Circuits - Power Factor and Wattless Current

What happens to the current in an AC circuit containing only an inductor when the source frequency approaches zero?

**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. In a purely inductive circuit, X_L = ω L , and I = (V/X_L) . As frequency ( f ) approaches zero, ω = 2π f also approaches zero, making X_L very small. Thus, the current increases significantly, approaching a maximum limited only by resistance (which is zero in an ideal inductor). Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L -

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Power in AC Circuits - Power Factor and Wattless Current

In an AC circuit with a series LCR combination, why does the power dissipated depend only on the resistive component?

**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. Power dissipation in an AC circuit ( P = I² R cos Φ ) occurs only through resistance, as inductors and capacitors store and release energy without converting it to heat. The reactive components affect the current and phase, but only R dissipates power. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² +

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