Skip to content

#AC analysis

29 public questions tagged with this topic.

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 \( 200 \, \text{V} \) (rms) AC source is connected to a series LCR circuit with \( R = 20 \, \Omega \) at resonance. W

**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. At resonance, Z = R = 20 Ω . RMS current: I = (V/R) = (200/20) = 10 A . Power: P = I² R = 10² × 20 = 2000 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 2000

Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor

In an LCR circuit with \( R = 3 \, \Omega \), \( X_L = 8 \, \Omega \), \( X_C = 4 \, \Omega \), what is the power factor

**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. Impedance: Z = √(R² + (X_L - X_C)²) = √(3² + (8 - 4)²) = √(9 + 16) = 5 Ω . Power factor: cos Φ = (R/Z) = (3/5) = 0.6 . 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,

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

A \( 24 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the r

**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. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 24 × 10⁻⁶ F . X_C = (1/314 × 24 × 10⁻⁶) ≈ 132.6 Ω . RMS current: I = (V/X_C) = (230/132.6) ≈ 1.734 A . 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 > Resonance in LCR Circuit and Q-Factor

In an AC circuit with a resistor and capacitor in series, what happens to the total voltage across the components compar

**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. In an RC series circuit, the voltages across the resistor ( V_R ) and capacitor ( V_C ) are 90° out of phase. The total source voltage is the vector sum, V = √(V_R² + V_C²) , which equals the applied voltage, not the algebraic sum, due to the phase difference. Applying X_L = ωL, X_C = 1/ωC, Z

Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor

A \( 26 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the r

**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. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 26 × 10⁻⁶ F . X_C = (1/314 × 26 × 10⁻⁶) ≈ 122.4 Ω . RMS current: I = (V/X_C) = (230/122.4) ≈ 1.879 A . 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 > Resonance in LCR Circuit and Q-Factor

A series LCR circuit has \( R = 200 \, \Omega \), \( C = 15 \, \mu\text{F} \), and is connected to a \( 220 \, \text{V}

**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. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s , C = 15 × 10⁻⁶ F . X_C = (1/314 × 15 × 10⁻⁶) ≈ 212.3 Ω . No inductor, so X_L = 0 . Impedance: Z = √(R² + (X_C - X_L)²) = √(200² + 212.3²) ≈ 291.5 Ω . Applying X_L = ωL, X_C =

Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor

A series LCR circuit has \( L = 2.5 \, \text{H} \), \( C = 40 \, \mu\text{F} \). What is the resonant angular frequency?

**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. ω₀ = (1/√(L C)) . L = 2.5 H , C = 40 × 10⁻⁶ F . ω₀ = (1/√(2.5 × 40 × 10⁻⁶)) = (1/√(10⁻⁴)) = 100 rad/s . 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 100 rad/s, consistent with phasor analysis and resonance condition X_L = X_C.

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

A series LCR circuit with \( R = 50 \, \Omega \) is at resonance with a \( 250 \, \text{V} \) (rms) source. What is the

**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. At resonance, Z = R = 50 Ω . RMS current: I = (V/R) = (250/50) = 5 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 5 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

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

**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