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#series AC circuit

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Power in AC Circuits - Power Factor and Wattless Current

A \( 70 \, \Omega \) resistor and \( 14 \, \mu\text{F} \) capacitor are in series with a \( 210 \, \text{V} \), \( 50 \,

**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_C = (1/ω C) = (1/314 × 14 × 10⁻⁶) ≈ 227.5 Ω . Z = √(R² + X_C²) = √(70² + 227.5²) = √(4900 + 51756.25) ≈ 238.2 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 238.2 Ω, consistent with phasor analysis and resonance condition X_L = X_C.

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

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AC Through Resistor - Phasor and Power

In an AC circuit with a capacitor and resistor in series, how does the phase difference between voltage and current chan

**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 RC series circuit, the phase angle Φ = tan⁻¹ ( (X_C/R) ) , where X_C = (1/ω C) . Increasing capacitance decreases X_C , reducing the phase angle, meaning the current leads the voltage by a smaller angle. 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 Through Resistor - Phasor and Power

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