#inductor

Energy Stored in Inductor and Magnetic Energy

A coil with \( L = 0.35 \, \text{H} \) has its current increased from 0 to 5 A in 0.5 s. What is the energy stored?

**Energy stored in inductor** U =½ L I² (J), L inductance (H), I current (A), resides in magnetic field, energy density u = B²/(2μ₀) (J/m³), B field (T), μ₀=4π×10⁻⁷ H/m. For solenoid B=μ₀ n I, volume V= A l, U = (B²/2μ₀) V =½ (μ₀ n² A l) I² =½ L I², consistent. W = (1/2) L I² = (1/2) × 0.35 × (5)² = 0.175 × 25 = 4.375 J ≈ 4.38 J . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M =

Ref: NCERT > Physics Book > Electromagnetic Induction > Energy Stored in Inductor and Magnetic Energy

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Transformer, AC Generator and LC Oscillations

A \( 40 \, \text{mH} \) inductor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the r

**Transformer principle** alternating current in primary creates changing flux linking secondary, emf induced e = -N dΦ/dt, flux same through both, so V∝N. For N_p=550, N_s=1100, V_p=110 V rms, V_s= V_p×N_s/N_p=220 V, step-up factor 2, efficiency

Ref: NCERT > Physics Book > Alternating Currents > Transformer, AC Generator and LC Oscillations

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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

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 \( 220 \, \text{V} \) (rms), \( 50 \, \text{Hz} \) AC source is connected to a \( 44 \, \text{mH} \) inductor. Calcula

**Condition for resistive behavior** X_L = X_C, so ωL=1/ωC, ω²=1/LC, this is resonance, impedance purely resistive Z=R, current maximum, power maximum P= V²/R, inductive and capacitive reactances equal magnitude opposite sign cancel, circuit appears resistive. Inductive reactance: X_L = ω L , where ω = 2π f . Given: f = 50 Hz , L = 44 mH = 0.044 H . ω = 2 × 3.14 × 50 = 314 rad/s . X_L = 314 × 0.044 = 13.816 Ω ≈ 13.82 Ω . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms

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 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

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

**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. X_L = ω L , ω = 2π × 60 = 376.8 rad/s . L = 60 × 10⁻³ H . X_L = 376.8 × 0.06 = 22.61 Ω . RMS current: I = (V/X_L) = (110/22.61) ≈ 4.87 A . Peak current: i_m = √(2) I = 1.414 × 4.87 ≈ 6.88 A . Applying X_L = ωL,

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

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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

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

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AC Through Capacitor - Capacitive Reactance

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

**AC through capacitor** voltage lags current by 90°, Q= C V, I= dQ/dt = C dV/dt, V(t)=V_peak sin ωt, I(t)=I_peak sin(ωt+90°), average power zero because energy stored in electric field ½ C V² returned each cycle, capacitor blocks DC but passes AC, X_C decreases with f. X_L = ω L , ω = 2π × 50 = 314 rad/s . L = 50 × 10⁻³ H . X_L = 314 × 0.05 = 15.7 Ω . RMS current: I = (V/X_L) = (220/15.7) ≈ 14.01 A . Peak current: i_m = √(2) I = 1.414 × 14.01 ≈ 19.81 A . Applying X_L = ωL,

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

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AC Through Capacitor - Capacitive Reactance

A \( 30 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the rms

**Capacitor average power** zero over complete cycle because P=V I =½ V_peak I_peak sin2ωt average zero, energy stored in field, not dissipated, unlike resistor. For 15 μF, 60 Hz, X_C=176.8 Ω, V_rms=110 V, I_rms=0.622 A, illustrating lower C higher X_C. X_L = ω L , ω = 2π × 50 = 314 rad/s . L = 30 × 10⁻³ H . X_L = 314 × 0.03 = 9.42 Ω . RMS current: I = (V/X_L) = (220/9.42) ≈ 23.35 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 =

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

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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

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LCR Series Circuit - Impedance and Phasor Diagram

In a series LCR circuit, why does resonance not occur if either the inductor or capacitor is absent?

**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. Resonance in a series LCR circuit requires both an inductor and capacitor to create a condition where X_L = X_C , canceling the reactive components. Without either, there’s no opposing reactance to balance, preventing resonance. 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 > LCR Series Circuit - Impedance and Phasor Diagram

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Resonance in LCR Circuit and Q-Factor

Why does the voltage across the inductor in a series LCR circuit equal the voltage across the capacitor at resonance, de

**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, X_L = X_C , and since V_L = I X_L and V_C = I X_C with the same current I , their magnitudes are equal. However, they are 180° out of phase, so their vector sum is zero, not their magnitudes. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms =

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

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