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#power dissipation

27 public questions tagged with this topic.

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

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

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Resonance in LCR Circuit and Q-Factor

A \( 200 \, \text{V} \) (rms) AC source is connected to a series LCR circuit with \( R = 20 \, \Omega \) at resonance. W

**Resonance in LCR** occurs when X_L = X_C, ω₀ =1/√(LC), f₀=1/(2π√(LC)), impedance Z=R minimum, current maximum I₀=V/R, circuit behaves as if only resistance because inductive and capacitive voltages equal opposite cancel, net reactance zero, phase φ=0°, power factor 1, current in phase with voltage. At resonance, Z = R = 20 Ω . RMS current: I = (V/R) = (200/20) = 10 A . Power: P = I² R = 10² × 20 = 2000 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 2000

Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor

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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 +

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

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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit with \( R = 50 \, \Omega \), \( X_L = 70 \, \Omega \), \( X_C = 30 \, \Omega \) has a \( 200 \, \te

**LCR series impedance** Z = √(R² + (X_L - X_C)²), R resistance (Ω), X_L=ωL, X_C=1/ωC, phase angle φ = tan⁻¹((X_L-X_C)/R), current I_rms = V_rms/Z, voltage across R in phase with I, across L leads by 90°, across C lags by 90°, phasor diagram vector sum V = √(V_R² + (V_L - V_C)²). Z = √(R² + (X_L - X_C)²) = √(50² + (70 - 30)²) = √(2500 + 1600) = √(4100) ≈ 64 Ω . RMS current: I = (V/Z) = (200/64) ≈ 3.125 A . Power: P = I² R = (3.125)² × 50 ≈ 488.28 W . Applying X_L = ωL, X_C =

Ref: NCERT > Physics Book > Alternating Currents > LCR Series Circuit - Impedance and Phasor Diagram

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Power in AC Circuits - Power Factor and Wattless Current

In an AC circuit with a pure inductor, why is the average power dissipated zero over a complete cycle?

**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. In a pure inductor, the current lags the voltage by 90°. The instantaneous power oscillates between positive (energy stored) and negative (energy returned), averaging to zero over a cycle because the inductor does not dissipate energy as heat but stores and releases it. Applying X_L = ωL, X_C = 1/ωC, Z

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

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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit with \( R = 110 \, \Omega \), \( X_L = 140 \, \Omega \), \( X_C = 80 \, \Omega \) has a \( 330 \, \

**Impedance behavior at high frequencies** X_L=ωL dominates ∝ f, X_C=1/ωC →0, so Z≈√(R²+X_L²)≈X_L large, current small, circuit inductive, φ→90°, at low frequencies X_C large, Z≈X_C, capacitive, φ→-90°, at intermediate resonance Z minimal =R. Z = √(R² + (X_L - X_C)²) = √(110² + (140 - 80)²) = √(12100 + 3600) = √(15700) ≈ 125.3 Ω . RMS current: I = (V/Z) = (330/125.3) ≈ 2.634 A . Power: P = I² R = (2.634)² × 110 ≈ 763.2 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,

Ref: NCERT > Physics Book > Alternating Currents > LCR Series Circuit - Impedance and Phasor Diagram

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AC Through Capacitor - Capacitive Reactance

A series LCR circuit with \( R = 60 \, \Omega \), \( X_L = 80 \, \Omega \), \( X_C = 20 \, \Omega \) has a \( 180 \, \te

**Capacitive reactance** X_C =1/(ω C)=1/(2π f C) (Ω), C capacitance (F), current leads voltage by 90°, I_rms = V_rms/X_C = V_rms ω C, I_peak = V_peak ω C, impedance Z = X_C for pure C. For 45 μF, 60 Hz, X_C=1/(2π×60×45×10⁻⁶)=58.9 Ω, V_rms=110 V, I_rms=1.867 A, I_peak=2.64 A. Z = √(R² + (X_L - X_C)²) = √(60² + (80 - 20)²) = √(3600 + 3600) = √(7200) ≈ 84.85 Ω . RMS current: I = (V/Z) = (180/84.85) ≈ 2.12 A . Power: P = I² R = (2.12)² × 60 ≈ 269.66 W . Applying X_L = ωL, X_C = 1/ωC, Z = √(R²

Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance

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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 =

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 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 =

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

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Power in AC Circuits - Power Factor and Wattless Current

What is the condition for maximum power dissipation in a series LCR circuit?

**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. Maximum power dissipation in a series LCR circuit occurs at resonance, where X_L = X_C , making the impedance equal to the resistance ( Z = R ). At this point, the power factor ( cos Φ ) is 1, and power ( P = I² R ) is maximized. Applying X_L

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

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Resonance in LCR Circuit and Q-Factor

In a series LCR circuit at resonance, the power dissipated is \( 4800 \, \text{W} \) with \( R = 3 \, \Omega \). What is

**At resonance** V_L = I X_L = I X_C = V_C, may be larger than source voltage Q times, Q-factor = ω₀ L/R =1/(ω₀ C R)= V_L/V = V_C/V, measures sharpness, higher Q sharper resonance, bandwidth Δω = R/L = ω₀/Q, resonant frequency independent of R. At resonance, Z = R , and power P = I² R . 4800 = I² × 3 . I² = (4800/3) = 1600 . I = √(1600) = 40 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

Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor

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Power in AC Circuits - Power Factor and Wattless Current

What is the average power dissipated in a purely capacitive circuit over one complete cycle?

**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. Power: p_C = i v = i_m v_m cos (ω t) sin (ω t) = (i_m v_m/2) sin (2ω t) . Average over a cycle: P_C = (i_m v_m/2) langle sin (2ω t) rangle = 0 , since langle sin (2ω t) rangle = 0 . Applying X_L = ωL, X_C =

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

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