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#220 V

2 public questions tagged with this topic.

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 Inductor - Inductive Reactance

A \( 22 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is th

**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_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 22 × 10⁻⁶ F . X_C = (1/314 × 22 × 10⁻⁶) ≈ 144.7 Ω . RMS current: I = (V/X_C) = (220/144.7) ≈ 1.52 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 1.52

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

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