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#inductors

9 public questions tagged with this topic.

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,

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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

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AC Through Inductor - Inductive Reactance

A \( 95 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the i

**AC through inductor** voltage leads current by 90°, V = L dI/dt, V(t)=V_peak sin(ωt+90°), I(t)=I_peak sin ωt, instantaneous power P= V I =½ V_peak I_peak sin2ωt, average zero over cycle because energy stored in magnetic field ½ L I² returned to source each quarter cycle, no net dissipation. X_L = ω L , ω = 2π f . f = 50 Hz , L = 95 × 10⁻³ H . ω = 2 × 3.14 × 50 = 314 rad/s . X_L = 314 × 0.095 = 29.83 Ω . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms

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AC Through Inductor - Inductive Reactance

A \( 90 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the i

**Inductive reactance calculation** for 60 mH, 50 Hz, X_L=2π×50×0.06=18.85 Ω, V_rms=220 V, I_rms=220/18.85=11.67 A, for 70 mH, 50 Hz, X_L=21.99 Ω, I=110/21.99=5 A, showing X_L ∝ f L, higher f or L increases opposition. X_L = ω L , ω = 2π f . f = 50 Hz , L = 90 × 10⁻³ H . ω = 2 × 3.14 × 50 = 314 rad/s . X_L = 314 × 0.09 = 28.26 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 25 Ω,

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AC Through Inductor - Inductive Reactance

A \( 90 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is the r

**Inductive reactance calculation** for 60 mH, 50 Hz, X_L=2π×50×0.06=18.85 Ω, V_rms=220 V, I_rms=220/18.85=11.67 A, for 70 mH, 50 Hz, X_L=21.99 Ω, I=110/21.99=5 A, showing X_L ∝ f L, higher f or L increases opposition. X_L = ω L , ω = 2π × 60 = 376.8 rad/s . L = 90 × 10⁻³ H . X_L = 376.8 × 0.09 = 33.91 Ω . RMS current: I = (V/X_L) = (220/33.91) ≈ 6.49 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 6.49 A, consistent

Ref: NCERT > Physics Book > Alternating Currents > AC Through Inductor - Inductive Reactance

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AC Through Inductor - Inductive Reactance

A \( 55 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the peak

**Inductive reactance** X_L = ω L =2π f L (Ω), L inductance (H), f frequency (Hz), ω=2πf angular frequency (rad/s), opposes AC, impedance Z = X_L for pure L, current lags voltage by 90°, I_rms = V_rms/X_L, I_peak = V_peak/X_L. For 85 mH, 50 Hz, X_L=2π×50×0.085=26.7 Ω, V_rms=230 V, I_rms=8.61 A, I_peak=12.18 A. X_L = ω L , ω = 2π × 60 = 376.8 rad/s . L = 55 × 10⁻³ H . X_L = 376.8 × 0.055 = 20.72 Ω . RMS current: I = (V/X_L) = (110/20.72) ≈ 5.31 A . Peak current: i_m = √(2) I = 1.414 × 5.31 ≈

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AC Through Inductor - Inductive Reactance

A \( 120 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the

**AC through inductor** voltage leads current by 90°, V = L dI/dt, V(t)=V_peak sin(ωt+90°), I(t)=I_peak sin ωt, instantaneous power P= V I =½ V_peak I_peak sin2ωt, average zero over cycle because energy stored in magnetic field ½ L I² returned to source each quarter cycle, no net dissipation. X_L = ω L , ω = 2π f . f = 50 Hz , L = 120 × 10⁻³ H . ω = 2 × 3.14 × 50 = 314 rad/s . X_L = 314 × 0.12 = 37.68 Ω . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms

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