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Alternating Currents

Latest questions in this category.

237 questions

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

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

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

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

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

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

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

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 \( 23 \, \mu\text{F} \) 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