For a series RLC circuit driven by voltage v = v m sinωt, the current is given by i = i m sin(ωt + φ) and the power fact
Power factor cosφ=R/Z, φ phase between V and I, P=VI cosφ.
Ref: NCERT Physics Class 12 Chapter 4 Moving Charges and Magnetism
#AC power
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A \( 180 \, \text{V} \) (rms) AC source supplies a \( 90 \, \Omega \) 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) = (180/90) = 2 A . Average power: P = I² R = 2² × 90 = 360 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
A \( 150 \, \text{V} \) (rms) AC source supplies a \( 75 \, \Omega \) resistor. What is the average power consumed?
**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. RMS current: I = (V/R) = (150/75) = 2 A . Average power: P = I² R = 2² × 75 = 4 × 75 = 300 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 300
Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance
A series LCR circuit with \( R = 60 \, \Omega \), \( X_L = 80 \, \Omega \), \( X_C = 20 \, \Omega \) has a \( 180 \, \te
**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. Z = √(R² + (X_L - X_C)²) = √(60² + (80 - 20)²) = √(3600 + 3600) = √(7200) ≈ 84.85 Ω . RMS current: I = (V/Z) = (180/84.85) ≈ 2.12 A . Power: P = I² R = (2.12)² × 60 ≈ 269.66 W . Applying X_L = ωL, X_C = 1/ωC, Z = √(R²
Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance
A \( 30 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the r
**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_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 30 × 10⁻⁶ F . X_C = (1/314 × 30 × 10⁻⁶) ≈ 106.1 Ω . RMS current: I = (V/X_C) = (230/106.1) ≈ 2.17 A . Applying X_L = ωL, X_C = 1/ωC, Z =
Ref: NCERT > Physics Book > Alternating Currents > AC Through Inductor - Inductive Reactance
A transformer has \( N_p = 500 \), \( N_s = 250 \). If \( V_s = 120 \, \text{V} \) (rms), what is the primary voltage?
**Transformer** works on mutual induction, V_s/V_p = N_s/N_p, I_s/I_p = N_p/N_s for ideal (power conserved V_p I_p = V_s I_s), step-up N_s>N_p V_s>V_p I_s
Ref: NCERT > Physics Book > Alternating Currents > Transformer, AC Generator and LC Oscillations
What is the average power dissipated in a purely capacitive circuit over one complete cycle?
**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. Power: p_C = i v = i_m v_m cos (ω t) sin (ω t) = (i_m v_m/2) sin (2ω t) . Average over a cycle: P_C = (i_m v_m/2) langle sin (2ω t) rangle = 0 , since langle sin (2ω t) rangle = 0 . Applying X_L = ωL, X_C =
Ref: NCERT > Physics Book > Alternating Currents > Power in AC Circuits - Power Factor and Wattless Current
A series LCR circuit with \( R = 80 \, \Omega \), \( X_L = 100 \, \Omega \), \( X_C = 40 \, \Omega \) has a \( 240 \, \t
**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. Z = √(R² + (X_L - X_C)²) = √(80² + (100 - 40)²) = √(6400 + 3600) = √(10000) = 100 Ω . RMS current: I = (V/Z) = (240/100) = 2.4 A . Power: P = I² R = (2.4)² × 80 = 5.76 × 80 = 460.8 W . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms
Ref: NCERT > Physics Book > Alternating Currents > AC Fundamentals - RMS, Average and Peak Values
