Skip to content
Login Register

#inductive reactance

22 public questions tagged with this topic.

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

View Details
LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( R = 120 \, \Omega \), \( X_L = 100 \, \Omega \), \( X_C = 80 \, \Omega \). What is the imped

**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) . Z = √(120² + (100 - 80)²) = √(14400 + 400) = √(14800) ≈ 121.66 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 121.66

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

View Details
LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( R = 15 \, \Omega \), \( X_L = 30 \, \Omega \), \( X_C = 45 \, \Omega \). What is the phase a

**LCR example** R=100 Ω, X_L=130 Ω, X_C=70 Ω, X_L-X_C=60 Ω, Z=√(100²+60²)=116.6 Ω, V_rms=300 V, I_rms=2.573 A, power P= I_rms² R =662 W? Actually P= V_rms I_rms cos φ, cos φ=R/Z=0.857, P=300×2.573×0.857=661.6 W, illustrating power factor. tan Φ = (X_C - X_L/R) = (45 - 30/15) = 1 . Φ = tan⁻¹(1) = 45° . Since X_C > X_L , current leads voltage. 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 45°, consistent with phasor analysis and resonance condition X_L = X_C.

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

View Details
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

View Details
AC Through Resistor - Phasor and Power

In an AC circuit with a resistor and inductor in series, what determines the magnitude of the phase difference between v

**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. In an RL series circuit, the phase angle Φ = tan⁻¹ ( (X_L/R) ) . The magnitude of this angle depends on the ratio of inductive reactance ( X_L = ω L ) to resistance ( R ), as it reflects the relative contributions of inductance and resistance. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z,

Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power

View Details
AC Through Resistor - Phasor and Power

In an AC circuit with a series combination of resistor and inductor, what happens to the impedance if the frequency decr

**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 RL series circuit, impedance Z = √(R² + X_L²) , where X_L = ω L . Decreasing frequency reduces ω , lowering X_L , which decreases the impedance since R 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 decreases, consistent with phasor analysis and resonance cond

Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power

View Details
AC Through Capacitor - Capacitive Reactance

A \( 75 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is the r

**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. X_L = ω L , ω = 2π × 60 = 376.8 rad/s . L = 75 × 10⁻³ H . X_L = 376.8 × 0.075 = 28.26 Ω . RMS current: I = (V/X_L) = (110/28.26) ≈ 3.89 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms =

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

View Details
LCR Series Circuit - Impedance and Phasor Diagram

Why does the impedance of an AC circuit with only an inductor increase with frequency?

**LCR example** R=100 Ω, X_L=130 Ω, X_C=70 Ω, X_L-X_C=60 Ω, Z=√(100²+60²)=116.6 Ω, V_rms=300 V, I_rms=2.573 A, power P= I_rms² R =662 W? Actually P= V_rms I_rms cos φ, cos φ=R/Z=0.857, P=300×2.573×0.857=661.6 W, illustrating power factor. The impedance in a purely inductive circuit is the inductive reactance ( X_L = ω L ), where ω = 2π f . As frequency ( f ) increases, ω increases linearly, causing X_L (and thus impedance) to increase proportionally. 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 Because indu

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

View Details
AC Through Resistor - Phasor and Power

A \( 110 \, \text{mH} \) inductor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the

**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_L = ω L , ω = 2π f . f = 50 Hz , L = 110 × 10⁻³ H . ω = 2 × 3.14 × 50 = 314 rad/s . X_L = 314 × 0.11 = 34.54 Ω . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L -

Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power

View Details
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,

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

View Details
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

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

View Details
LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit with \( R = 50 \, \Omega \), \( X_L = 30 \, \Omega \), \( X_C = 20 \, \Omega \) is connected to an

**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. Impedance: Z = √(R² + (X_L - X_C)²) . Z = √(50² + (30 - 20)²) = √(2500 + 100) = √(2600) ≈ 51 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 50 Ω, consistent with phasor analysis and resonance condition X_L = X_C.

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

View Details