28 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
**AC generator** converts mechanical to electrical, emf e = N B A ω sin ωt, maximum e₀ = N B A ω, frequency = rotation frequency, transformer cannot work on DC because steady flux no induction, LC oscillations energy swaps between ½ L I² and ½ Q²/C at ω₀=1/√(LC). Z = √(R² + (X_L - X_C)²) . Z = √(70² + (40 - 20)²) = √(4900 + 400) = √(5300) ≈ 72.8 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 72.8 Ω, consistent with
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_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
**Resonance in LCR** occurs when X_L = X_C, ω₀ =1/√(LC), f₀=1/(2π√(LC)), impedance Z=R minimum, current maximum I₀=V/R, circuit behaves as if only resistance because inductive and capacitive voltages equal opposite cancel, net reactance zero, phase φ=0°, power factor 1, current in phase with voltage. Z = √(R² + (X_L - X_C)²) . Z = √(60² + (50 - 30)²) = √(3600 + 400) = √(4000) ≈ 63.25 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 62 Ω, consistent with phasor analysis and resonance condition X_L = X_C.
Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor
**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_C = (1/ω C) = (1/314 × 15 × 10⁻⁶) ≈ 212.3 Ω . Z = √(R² + X_C²) = √(60² + 212.3²) = √(3600 + 45071.29) ≈ 220.8 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 220.8 Ω, consistent with phasor analysis and resonance condition X_L = X_C.
Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power
**LCR example** R=100 Ω, X_L=130 Ω, X_C=70 Ω, X_L-X_C=60 Ω, Z=√(100²+60²)=116.6 Ω, V_rms=300 V, I_rms=2.573 A, power P= I_rms² R =662 W? Actually P= V_rms I_rms cos φ, cos φ=R/Z=0.857, P=300×2.573×0.857=661.6 W, illustrating power factor. Z = √(R² + (X_L - X_C)²) . Z = √(80² + (60 - 40)²) = √(6400 + 400) = √(6800) ≈ 82.46 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 82.46 Ω, consistent with phasor analysis and resonance condition X_L = X_C.
Ref: NCERT > Physics Book > Alternating Currents > LCR Series Circuit - Impedance and Phasor Diagram
**LCR series impedance** Z = √(R² + (X_L - X_C)²), R resistance (Ω), X_L=ωL, X_C=1/ωC, phase angle φ = tan⁻¹((X_L-X_C)/R), current I_rms = V_rms/Z, voltage across R in phase with I, across L leads by 90°, across C lags by 90°, phasor diagram vector sum V = √(V_R² + (V_L - V_C)²). Z = √(R² + (X_L - X_C)²) . Z = √(120² + (100 - 80)²) = √(14400 + 400) = √(14800) ≈ 121.66 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 121.66
Ref: NCERT > Physics Book > Alternating Currents > LCR Series Circuit - Impedance and Phasor Diagram
**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. X_C = (1/ω C) = (1/314 × 20 × 10⁻⁶) ≈ 159.2 Ω . Z = √(R² + X_C²) = √(50² + 159.2²) = √(2500 + 25344.64) ≈ 166.6 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 =
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. In an RL series circuit, impedance Z = √(R² + X_L²) , where X_L = ω L . Decreasing frequency reduces ω , lowering X_L , which decreases the impedance since R remains constant. 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 It decreases, consistent with phasor analysis and resonance cond
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_C = (1/ω C) = (1/314 × 18 × 10⁻⁶) ≈ 176.8 Ω . Z = √(R² + X_C²) = √(45² + 176.8²) = √(2026 + 31258.24) ≈ 182.4 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 182.4 Ω, consistent with phasor analysis and resonance condition X_L = X_C.
Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power
**Average power in LCR** P_avg = V_rms I_rms cos φ, cos φ = R/Z power factor, φ phase between V and I, tan φ = (X_L - X_C)/R. For R=80 Ω, X_L=100 Ω, X_C=40 Ω, X_L-X_C=60 Ω, Z=√(80²+60²)=100 Ω, cos φ=0.8, V_rms=240 V, I_rms=2.4 A, P=240×2.4×0.8=460.8 W, only R dissipates. Z = √(R² + (X_L - X_C)²) = √(10² + (15 - 5)²) = √(100 + 100) = 14.14 Ω . Power factor: cos Φ = (R/Z) = (10/14.14) ≈ 0.707 . 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
Ref: NCERT > Physics Book > Alternating Currents > Power in AC Circuits - Power Factor and Wattless Current
**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) = √(90² + (120 - 60)²) = √(8100 + 3600) = √(11700) ≈ 108.17 Ω . RMS current: I = (V/Z) = (270/108.17) ≈ 2.496 A . Power: P = I² R = (2.496)² × 90 ≈ 560.6 W . 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 Fundamentals - RMS, Average and Peak Values
