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#transformer

17 public questions tagged with this topic.

Rectifiers, Filters and Applications

The purpose of a centre-tap transformer in a full-wave rectifier is to:

**Rectifier** converts AC to DC using diode unilateral conduction, half-wave conducts only during positive half-cycle when forward biased, output pulsating DC with frequency equal to input 50 Hz, full-wave with centre-tap uses two diodes conducting alternate half-cycles, output frequency 100 Hz for 50 Hz input, 120 Hz for 60 Hz input, output voltage closer to peak with capacitor filter. The centre-tap transformer splits the secondary voltage into two equal halves, allowing two diodes to rectify alternate half-cycles, producing a full-wave output. Substituting values gives Split the AC input, w

Ref: NCERT > Physics Book > Electronic Devices > Rectifiers, Filters and Applications

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Transformer, AC Generator and LC Oscillations

A transformer has \( N_p = 600 \), \( N_s = 300 \). If \( V_p = 240 \, \text{V} \) (rms), what is the secondary voltage?

**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 transformer has \( N_p = 600 \), \( N_s = 1200 \). If \( V_s = 480 \, \text{V} \) (rms), what is the primary voltage?

**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). (V_s/V_p) = (N_s/N_p) . V_p = V_s × (N_p/N_s) = 480 × (600/1200) = 240 V . 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 240 V, consistent with phasor analysis and resonance condition X_L = X_C.

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

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AC Through Resistor - Phasor and Power

What is the effect on an ideal transformer’s secondary voltage if the number of turns in the secondary coil is halved?

**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. In an ideal transformer, (V_s/V_p) = (N_s/N_p) . If the number of secondary turns ( N_s ) is halved, the secondary voltage ( V_s ) becomes half its original value, assuming primary voltage ( V_p ) remains constant. 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 It halves, consistent wit

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

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Transformer, AC Generator and LC Oscillations

A transformer has \( N_p = 250 \), \( N_s = 500 \). If \( V_s = 440 \, \text{V} \) (rms), what is the primary voltage?

**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). (V_s/V_p) = (N_s/N_p) . V_p = V_s × (N_p/N_s) = 440 × (250/500) = 220 V . 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 220 V, consistent with phasor analysis and resonance condition X_L = X_C.

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

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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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Transformer, AC Generator and LC Oscillations

A transformer has \( N_p = 800 \), \( N_s = 400 \). If \( V_s = 110 \, \text{V} \) (rms), what is the primary voltage?

**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). (V_s/V_p) = (N_s/N_p) . V_p = V_s × (N_p/N_s) = 110 × (800/400) = 220 V . 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 220 V, consistent with phasor analysis and resonance condition X_L = X_C.

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

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