26 public questions tagged with this topic.
**Transformer principle** alternating current in primary creates changing flux linking secondary, emf induced e = -N dΦ/dt, flux same through both, so V∝N. For N_p=550, N_s=1100, V_p=110 V rms, V_s= V_p×N_s/N_p=220 V, step-up factor 2, efficiency
Ref: NCERT > Physics Book > Alternating Currents > Transformer, AC Generator and LC Oscillations
**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_L = ω L , ω = 2π × 50 = 314 rad/s . L = 45 × 10⁻³ H . X_L = 314 × 0.045 = 14.13 Ω . RMS current: I = (V/X_L) = (230/14.13) ≈ 16.28 A . Peak current: i_m = √(2) I = 1.414 × 16.28 ≈ 23.02 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R²
Ref: NCERT > Physics Book > Alternating Currents > Power in AC Circuits - Power Factor and Wattless Current
**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. Capacitive reactance: X_C = (1/ω C) , where ω = 2π f . f = 60 Hz , C = 60 × 10⁻⁶ F . ω = 2 × 3.14 × 60 = 376.8 rad/s . X_C = (1/376.8 × 60 × 10⁻⁶) = 44.24 Ω . RMS current: I = (V/X_C)
Ref: NCERT > Physics Book > Alternating Currents > Power in AC Circuits - Power Factor and Wattless Current
**At resonance** V_L = I X_L = I X_C = V_C, may be larger than source voltage Q times, Q-factor = ω₀ L/R =1/(ω₀ C R)= V_L/V = V_C/V, measures sharpness, higher Q sharper resonance, bandwidth Δω = R/L = ω₀/Q, resonant frequency independent of R. X_L = ω L , ω = 2π × 60 = 376.8 rad/s . L = 60 × 10⁻³ H . X_L = 376.8 × 0.06 = 22.61 Ω . RMS current: I = (V/X_L) = (110/22.61) ≈ 4.87 A . Peak current: i_m = √(2) I = 1.414 × 4.87 ≈ 6.88 A . Applying X_L = ωL,
Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor
**At resonance** V_L = I X_L = I X_C = V_C, may be larger than source voltage Q times, Q-factor = ω₀ L/R =1/(ω₀ C R)= V_L/V = V_C/V, measures sharpness, higher Q sharper resonance, bandwidth Δω = R/L = ω₀/Q, resonant frequency independent of R. RMS current: I = (V/R) = (200/100) = 2 A . Peak current: i_m = √(2) I = 1.414 × 2 ≈ 2.828 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 2.828 A, consistent with phasor analysis and resonance condition X_L = X_C.
Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor
**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. RMS current: I = (V/R) = (240/120) = 2 A . Peak current: i_m = √(2) I = 1.414 × 2 ≈ 2.828 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 2.828 A, consistent with phasor analysis and resonance condition X_L = X_C.
Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power
**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
**AC through resistor** voltage and current in phase, φ=0°, I = V/R instantaneously, I(t)=I_peak sin ωt, V(t)=V_peak sin ωt, phasor diagram V and I same direction, power instantaneous P = V I = V_peak I_peak sin² ωt, average P_avg = V_rms I_rms = V_rms²/R, always positive, energy dissipated as heat. RMS current: I = (V/R) = (190/95) = 2 A . Peak current: i_m = √(2) I = 1.414 × 2 ≈ 2.828 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 2.828 A, consistent
Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power
**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. X_L = ω L , ω = 2π × 50 = 314 rad/s . L = 50 × 10⁻³ H . X_L = 314 × 0.05 = 15.7 Ω . RMS current: I = (V/X_L) = (220/15.7) ≈ 14.01 A . Peak current: i_m = √(2) I = 1.414 × 14.01 ≈ 19.81 A . Applying X_L = ωL,
Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance
**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. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 25 × 10⁻⁶ F . X_C = (1/314 × 25 × 10⁻⁶) ≈ 127.4 Ω . RMS current: I = (V/X_C) = (230/127.4) ≈ 1.805 A . Peak current: i_m = √(2) I = 1.414 × 1.805 ≈ 2.55 A .
Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance
**Capacitor average power** zero over complete cycle because P=V I =½ V_peak I_peak sin2ωt average zero, energy stored in field, not dissipated, unlike resistor. For 15 μF, 60 Hz, X_C=176.8 Ω, V_rms=110 V, I_rms=0.622 A, illustrating lower C higher X_C. X_C = (1/ω C) , ω = 2π × 60 = 376.8 rad/s . C = 20 × 10⁻⁶ F . X_C = (1/376.8 × 20 × 10⁻⁶) ≈ 132.7 Ω . RMS current: I = (V/X_C) = (110/132.7) ≈ 0.829 A . Peak current: i_m = √(2) I = 1.414 × 0.829 ≈ 1.17 A . Applying X_L = ωL, X_C = 1/ωC, Z
Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance
**RMS value** I_rms = I_peak/√2, V_rms = V_peak/√2 for sinusoidal AC, significance rms gives equivalent DC value producing same heating power P = I_rms² R, average power over cycle, instruments measure rms, average over full cycle zero, half-cycle average 2 I_peak/π, peak = √2 rms. RMS current: I = (V/R) = (180/90) = 2 A . Peak current: i_m = √(2) I = 1.414 × 2 ≈ 2.828 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 2.828 A, consistent with phasor analysis and resonance condition X_L = X_C.
Ref: NCERT > Physics Book > Alternating Currents > AC Fundamentals - RMS, Average and Peak Values
