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#electrical engineering

139 public questions tagged with this topic.

AC Generator, Back EMF and Time Duration of EMF

In an AC generator, slip rings are used instead of a split ring commutator. What is the primary reason for this design c

**Circular loop deformed into straight wire** in field B=0.12 T radius 16 cm area πr²=0.0804 m² flux 0.00965 Wb drops to zero in 0.6 s e=0.0161 V, illustrating flux change due to area change induces emf, even without B change, area deformation changes Φ = B A cosθ. Slip rings allow continuous contact with the coil, preserving the alternating nature of the emf, unlike a split ring commutator which rectifies it to DC. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt) and

Ref: NCERT > Physics Book > Electromagnetic Induction > AC Generator, Back EMF and Time Duration of EMF

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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

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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Resonance in LCR Circuit and Q-Factor

A \( 220 \, \text{V} \) (rms), \( 50 \, \text{Hz} \) AC source is connected to a \( 44 \, \text{mH} \) inductor. Calcula

**Condition for resistive behavior** X_L = X_C, so ωL=1/ωC, ω²=1/LC, this is resonance, impedance purely resistive Z=R, current maximum, power maximum P= V²/R, inductive and capacitive reactances equal magnitude opposite sign cancel, circuit appears resistive. Inductive reactance: X_L = ω L , where ω = 2π f . Given: f = 50 Hz , L = 44 mH = 0.044 H . ω = 2 × 3.14 × 50 = 314 rad/s . X_L = 314 × 0.044 = 13.816 Ω ≈ 13.82 Ω . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms

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

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Resonance in LCR Circuit and Q-Factor

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

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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

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