#AC voltage

Power in AC Circuits - Power Factor and Wattless Current

A \( 165 \, \text{V} \) (rms) AC source supplies a \( 55 \, \Omega \) resistor. What is the average power consumed?

**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. RMS current: I = (V/R) = (165/55) = 3 A . Average power: P = I² R = 3² × 55 = 495 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, calculation gives

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

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AC Through Capacitor - Capacitive Reactance

An AC voltage \( v = 141.4 \sin (100\pi t) \) V is applied across a \( 50 \, \Omega \) resistor. What is the rms current

**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. Given: v = v_m sin ω t , where v_m = 141.4 V , R = 50 Ω . RMS voltage: V = (v_m/√(2)) = (141.4/√(2)) = 100 V . RMS current: I = (V/R) = (100/50) = 2 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P

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

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AC Through Resistor - Phasor and Power

A \( 60 \, \Omega \) resistor and \( 40 \, \mu\text{F} \) capacitor are in series with a \( 230 \, \text{V} \), \( 50 \,

**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. X_C = (1/ω C) = (1/314 × 40 × 10⁻⁶) ≈ 79.6 Ω . Z = √(R² + X_C²) = √(60² + 79.6²) = √(3600 + 6336.16) ≈ 99.8 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 99.8 Ω, consistent with phasor analysis and resonance condition X_L = X_C.

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

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AC Through Inductor - Inductive Reactance

A \( 150 \, \Omega \) resistor is connected to a \( 300 \, \text{V} \) (peak) AC supply. What is the rms current in the

**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. Peak voltage: v_m = 300 V . RMS voltage: V = (v_m/√(2)) = (300/1.414) ≈ 212.1 V . RMS current: I = (V/R) = (212.1/150) ≈ 1.414 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 1.414 A, consistent with phasor analysis and resonance condition X_L = X_C.

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

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AC Fundamentals - RMS, Average and Peak Values

A \( 226.3 \, \text{V} \) (peak) AC source is connected to a \( 80 \, \Omega \) resistor. What is the average power cons

**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. RMS voltage: V = (v_m/√(2)) = (226.3/1.414) ≈ 160 V . RMS current: I = (V/R) = (160/80) = 2 A . Average power: P = I² R = 2² × 80 = 320 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, calculation gives 320 W, consistent

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

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