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Alternating Currents

The Alternating Currents category gathers questions that test understanding of AC concepts. Topics include sinusoidal waveforms, frequency, RMS calculations, phase relationships, and how alternating voltage and current behave in circuits.

237 questions

Transformer, AC Generator and LC Oscillations

Why does a step-down transformer increase the current in the secondary coil compared to the primary coil?

**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). In a step-down transformer ( V_s < V_p , N_s < N_p ), power is conserved ( V_p I_p = V_s I_s ). Since V_s is lower, I_s must be higher than I_p ( I_s = I_p × (N_p/N_s) , where (N_p/N_s) > 1 ) to maintain the same power output. Applying X_L = ωL, X_C

Ref: NCERT > Physics Book > Alternating Currents > Transformer, AC Generator and LC Oscillations

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Transformer, AC Generator and LC Oscillations

What is the effect on the impedance of a series LCR circuit when the frequency is increased beyond the resonant frequenc

**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

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Transformer, AC Generator and LC Oscillations

A series LCR circuit has \( R = 70 \, \Omega \), \( X_L = 40 \, \Omega \), \( X_C = 20 \, \Omega \). What is the impedan

**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

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Transformer, AC Generator and LC Oscillations

A \( 40 \, \text{mH} \) inductor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the r

**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

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Transformer, AC Generator and LC Oscillations

A \( 120 \, \Omega \) resistor is connected to a \( 240 \, \text{V} \) (rms) AC source. What is the rms current?

**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). RMS current: I = (V/R) . Given: V = 240 V , R = 120 Ω . I = (240/120) = 2 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 A, consistent with phasor

Ref: NCERT > Physics Book > Alternating Currents > Transformer, AC Generator and LC Oscillations

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Transformer, AC Generator and LC Oscillations

A \( 35 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is th

**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

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Power in AC Circuits - Power Factor and Wattless Current

A \( 45 \, \text{mH} \) inductor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the peak

**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

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Power in AC Circuits - Power Factor and Wattless Current

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

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Power in AC Circuits - Power Factor and Wattless Current

A \( 23 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the r

**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. X_C = (1/ω C) , ω = 2π × 60 = 376.8 rad/s . C = 23 × 10⁻⁶ F . X_C = (1/376.8 × 23 × 10⁻⁶) ≈ 115.4 Ω . RMS current: I = (V/X_C) = (110/115.4) ≈ 0.953 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L

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

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