#alternating current

Rotational EMF and AC Generator

In an AC generator, the emf varies sinusoidally with time. What is the primary reason for this variation?

**Rotational emf** when coil area A rotates with angular speed ω in uniform field B, flux Φ = B A cos ωt, emf e = -N dΦ/dt = N B A ω sin ωt, maximum e₀ = N B A ω, frequency = ω/2π. For square side 22 cm area 0.0484 m² N=1 ω=14 rad/s B=0.15 T, e₀=1×0.15×0.0484×14=0.1016 V, sinusoidal. The emf varies sinusoidally because the coil rotates in a uniform magnetic field, causing the angle between the magnetic field and the coil’s area vector to change continuously, following a sine function. Using Φ = B A cosθ, e = -N dΦ/dt = -N A

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

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AC Through Resistor - Phasor and Power

Why does a transformer not work with direct current (DC)?

**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. A transformer relies on a changing magnetic flux to induce voltage in the secondary coil via mutual induction. DC provides a constant current, producing a steady magnetic field with no flux change, so no voltage is induced. 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 g

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

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AC Fundamentals - RMS, Average and Peak Values

A \( 50 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is th

**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. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 50 × 10⁻⁶ F . X_C = (1/314 × 50 × 10⁻⁶) ≈ 63.7 Ω . RMS current: I = (V/X_C) = (220/63.7) ≈ 3.45 A . Peak current: i_m = √(2) I = 1.414 × 3.45 ≈ 4.88 A . Applying X_L =

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

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AC Fundamentals - RMS, Average and Peak Values

A \( 198.1 \, \text{V} \) (peak) AC source is connected to a \( 70 \, \Omega \) resistor. What is the average power cons

**Peak current** I_peak = V_peak/R for resistor, I_rms = V_rms/R, V_peak = √2 V_rms, for 200 V rms, V_peak=282.8 V, I_peak=282.8/80=3.535 A, rms I=200/80=2.5 A, average over complete cycle zero because positive and negative halves cancel. RMS voltage: V = (v_m/√(2)) = (198.1/1.414) ≈ 140 V . RMS current: I = (V/R) = (140/70) = 2 A . Average power: P = I² R = 2² × 70 = 280 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 280 W, consistent with phasor analysis

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

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AC Fundamentals - RMS, Average and Peak Values

A \( 35 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is the r

**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. X_L = ω L , ω = 2π × 60 = 376.8 rad/s . L = 35 × 10⁻³ H . X_L = 376.8 × 0.035 = 13.19 Ω . RMS current: I = (V/X_L) = (110/13.19) ≈ 8.34 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,

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

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AC Fundamentals - RMS, Average and Peak Values

A \( 100 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the

**Peak current** I_peak = V_peak/R for resistor, I_rms = V_rms/R, V_peak = √2 V_rms, for 200 V rms, V_peak=282.8 V, I_peak=282.8/80=3.535 A, rms I=200/80=2.5 A, average over complete cycle zero because positive and negative halves cancel. X_L = ω L , ω = 2π f . f = 50 Hz , L = 100 × 10⁻³ H . ω = 2 × 3.14 × 50 = 314 rad/s . X_L = 314 × 0.1 = 31.4 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

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

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AC Fundamentals - RMS, Average and Peak Values

A series LCR circuit with \( R = 90 \, \Omega \), \( X_L = 120 \, \Omega \), \( X_C = 60 \, \Omega \) has a \( 270 \, \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) = √(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

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AC Fundamentals - RMS, Average and Peak Values

What is the primary reason AC voltage is preferred over DC voltage for long-distance power transmission?

**Peak current** I_peak = V_peak/R for resistor, I_rms = V_rms/R, V_peak = √2 V_rms, for 200 V rms, V_peak=282.8 V, I_peak=282.8/80=3.535 A, rms I=200/80=2.5 A, average over complete cycle zero because positive and negative halves cancel. AC voltage is preferred for long-distance power transmission because it can be easily stepped up or down using transformers. This allows for high-voltage transmission to reduce energy losses due to resistance, followed by stepping down to safer levels for distribution. 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 c

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

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AC Through Inductor - Inductive Reactance

A \( 95 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the i

**AC through inductor** voltage leads current by 90°, V = L dI/dt, V(t)=V_peak sin(ωt+90°), I(t)=I_peak sin ωt, instantaneous power P= V I =½ V_peak I_peak sin2ωt, average zero over cycle because energy stored in magnetic field ½ L I² returned to source each quarter cycle, no net dissipation. X_L = ω L , ω = 2π f . f = 50 Hz , L = 95 × 10⁻³ H . ω = 2 × 3.14 × 50 = 314 rad/s . X_L = 314 × 0.095 = 29.83 Ω . 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 Inductor - Inductive Reactance

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AC Through Inductor - Inductive Reactance

A \( 16 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is th

**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 = 16 × 10⁻⁶ F . X_C = (1/314 × 16 × 10⁻⁶) ≈ 199 Ω . RMS current: I = (V/X_C) = (230/199) ≈ 1.156 A . Peak current: i_m = √(2) I = 1.414 ×

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

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AC Through Capacitor - Capacitive Reactance

A \( 40 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the c

**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 × 50 = 314 rad/s . C = 40 × 10⁻⁶ F . X_C = (1/314 × 40 × 10⁻⁶) ≈ 79.6 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 = N_s/N_p, calculation gives 79.6 Ω, consistent with phasor analysis

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

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Power in AC Circuits - Power Factor and Wattless Current

A series LCR circuit has \( R = 40 \, \Omega \), \( X_L = 65 \, \Omega \), \( X_C = 25 \, \Omega \). What is the power f

**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. Z = √(R² + (X_L - X_C)²) = √(40² + (65 - 25)²) = √(1600 + 1600) = √(3200) ≈ 56.57 Ω . Power factor: cos Φ = (R/Z) = (40/56.57) ≈ 0.707 . 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 > Power in AC Circuits - Power Factor and Wattless Current

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