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CBSE Class 12 Physics · 11 questions · 26 marks
Household supply never sits still — it swings between positive and negative fifty times a second, and that swinging is what lets transformers step voltage up and down almost losslessly. This chapter builds the phasor toolkit for AC circuits, shows how resistors, inductors and capacitors each treat a changing current differently, and ends at resonance, where an LCR circuit rings loudest at one particular frequency.
An alternating current has a peak value of 14.14 A. Its rms value is nearest to:
Answer
10 A — for a sinusoidal wave, i_rms = i₀/√2 = 14.14/1.414 = 10 A. This is the steady value of direct current that would dissipate heat in a resistor at the same average rate, which is why AC meters are calibrated to read rms values.
A pure inductor of inductance 200 mH is connected to a 50 Hz AC source. Its reactance is closest to:
Answer
62.8 Ω — X_L = ωL = 2πνL = 2 × 3.14 × 50 × 0.200 = 62.8 Ω. Reactance grows with frequency for an inductor, which is why inductors block high-frequency AC more than low-frequency AC.
In a series LCR circuit driven at its resonant frequency, which statement is correct?
Answer
Impedance equals R and current is maximum — at resonance X_L = X_C, so they cancel in Z = √[R² + (X_L - X_C)²], leaving Z = R, its smallest possible value. Since i_rms = e_rms/Z, the current is then at its peak, and the circuit behaves as if it were purely resistive with current in phase with voltage.
A step-down transformer converts 220 V AC to 11 V AC. If the primary carries a current of 0.50 A, the secondary current is (assume an ideal transformer):
Answer
10 A — for an ideal transformer, power is conserved: e_p i_p = e_s i_s. So i_s = e_p i_p/e_s = (220 × 0.50)/11 = 110/11 = 10 A. Stepping voltage down by a factor of 20 steps the current up by the same factor.
Assertion (A): A capacitor blocks direct current completely but allows alternating current to pass. Reason (R): The capacitive reactance X_C = 1/ωC becomes infinite as the frequency ω approaches zero, and decreases as ω increases.
Answer
Both A and R are true and R is the correct explanation of A — DC corresponds to ω = 0, where X_C → ∞, so no steady current flows once the capacitor is charged. For AC, ω is finite, X_C is finite, and a continuously alternating current does flow to charge and discharge the plates each half cycle.
A resistor of 40 Ω is connected across an AC source of rms voltage 200 V. Find the rms current and the average power dissipated.
Answer
For a pure resistor, current is in phase with voltage, so i_rms = e_rms/R = 200/40 = 5.0 A. Average power, P = e_rms i_rms cosφ, and here φ = 0 so cosφ = 1. P = 200 × 5.0 × 1 = 1000 W = 1.0 kW. This equals e_rms²/R = (200)²/40 = 1000 W, a useful cross-check.
A capacitor of 25 μF is connected to a 230 V, 50 Hz AC supply. Calculate its capacitive reactance and the rms current drawn, and state the phase relation between current and voltage.
Answer
Angular frequency ω = 2πν = 2 × 3.14 × 50 = 314 rad/s. X_C = 1/ωC = 1/(314 × 25×10⁻⁶) = 1/(7.85×10⁻³) = 127.4 Ω. Current, i_rms = e_rms/X_C = 230/127.4 = 1.81 A. In a pure capacitor, the current leads the applied voltage by a phase angle of 90°, since the charge (and hence current) responds ahead of the instantaneous voltage across the plates.
Define the power factor of an AC circuit and state its value for a circuit containing only an ideal inductor. Explain why current through a pure inductor is called a wattless current.
Answer
The power factor is cosφ, where φ is the phase difference between the current and the applied voltage; it measures what fraction of the apparent power e_rms i_rms is actually converted into average real power, P = e_rms i_rms cosφ. For a pure inductor, current lags voltage by 90°, so cosφ = cos90° = 0. Since P = e_rms i_rms cosφ = 0, no net energy is dissipated over a full cycle even though current flows; energy is stored in the magnetic field during one quarter cycle and returned to the source in the next, hence the term wattless current.
Derive an expression for the impedance of a series LCR circuit connected to an AC source, using a phasor diagram. A series LCR circuit has R = 30 Ω, L = 0.20 H and C = 20 μF, connected to a 200 V, 50 Hz supply. Find the impedance and the rms current.
Answer
Derivation: Let the same current i = i₀ sin ωt flow through R, L and C in series. The voltage phasors are: V_R = i₀R in phase with current; V_L = i₀X_L leading current by 90°; V_C = i₀X_C lagging current by 90°. Since V_L and V_C are exactly opposite in phase (both perpendicular to current but on opposite sides), their resultant is (V_L - V_C) = i₀(X_L - X_C), still perpendicular to the current phasor. The applied voltage is the phasor sum of V_R and (V_L - V_C), which are mutually perpendicular: V₀² = (i₀R)² + [i₀(X_L - X_C)]² V₀ = i₀√[R² + (X_L - X_C)²] Defining impedance Z = V₀/i₀: Z = √[R² + (X_L - X_C)²]. The phase angle by which current lags (or leads) the voltage is tanφ = (X_L - X_C)/R. Numerical: ω = 2πν = 314 rad/s. X_L = ωL = 314 × 0.20 = 62.8 Ω. X_C = 1/ωC = 1/(314 × 20×10⁻⁶) = 1/(6.28×10⁻³) = 159.2 Ω. (X_L - X_C) = 62.8 - 159.2 = -96.4 Ω. Z = √[(30)² + (-96.4)²] = √[900 + 9293] = √10193 = 100.96 Ω ≈ 101 Ω. i_rms = e_rms/Z = 200/101 = 1.98 A ≈ 2.0 A. Since X_C > X_L, the circuit is capacitive overall and the current leads the voltage.
(a) Explain the principle, construction and working of a transformer, and derive the relation e_s/e_p = N_s/N_p. (b) A transformer is used to light a 100 W, 220 V lamp from a 4400 V transmission line. Find the turns ratio required and the current drawn from the line, assuming 100% efficiency.
Answer
(a) Principle: a transformer works on mutual induction — a changing current in the primary coil sets up a changing magnetic flux, which links the secondary coil and induces an emf in it. Construction: two coils of insulated wire, the primary with N_p turns and the secondary with N_s turns, are wound on a common laminated soft-iron core, which confines almost all the flux to link both coils equally. Working: an alternating emf e_p applied to the primary drives a current that creates a flux Φ in the core, essentially the same for each turn of both coils since the core carries the flux with negligible leakage. By Faraday's law, the induced emf per turn is the same in both coils: emf per turn = dΦ/dt. So e_p = N_p (dΦ/dt) and e_s = N_s (dΦ/dt). Dividing: e_s/e_p = N_s/N_p. For an ideal (100% efficient) transformer, input power equals output power: e_p i_p = e_s i_s, so N_s/N_p = i_p/i_s as well. (b) Here e_p = 4400 V (line, treated as primary), e_s = 220 V (lamp, secondary). Turns ratio N_s/N_p = e_s/e_p = 220/4400 = 1/20, i.e. N_p : N_s = 20 : 1 (a step-down transformer). Secondary (lamp) current: i_s = P/e_s = 100/220 = 0.455 A. By power conservation, i_p = i_s × (N_s/N_p) = 0.455 × (1/20) = 0.0227 A ≈ 22.7 mA drawn from the 4400 V line.
Read the passage and answer the questions that follow: A radio receiver's tuning circuit is a series LCR circuit whose capacitance can be varied by turning a dial, changing the resonant frequency so that the circuit responds strongly to one broadcast station and rejects others. In one such circuit the coil has inductance 2.0 mH and negligible resistance is idealised as R = 5.0 Ω, and the listener wants to tune in to a station broadcasting at 1000 kHz (1.0×10⁶ Hz). (a) Calculate the capacitance needed for resonance at this frequency. (b) At resonance, what is the impedance of the circuit? (c) If the source supplies an rms voltage of 10 mV at resonance, find the rms current in the circuit. (d) Explain, in terms of X_L and X_C, why signals at frequencies far from 1000 kHz produce very little current in this circuit.
Answer
(a) At resonance, ω₀ = 1/√(LC), so C = 1/(ω₀²L). ω₀ = 2πν = 2 × 3.14 × 1.0×10⁶ = 6.28×10⁶ rad/s. ω₀² = 3.94×10¹³. C = 1/(3.94×10¹³ × 2.0×10⁻³) = 1/(7.88×10¹⁰) = 1.27×10⁻¹¹ F ≈ 12.7 pF. (b) At resonance X_L = X_C, so they cancel and Z = R = 5.0 Ω, the minimum possible impedance for this circuit. (c) i_rms = e_rms/Z = 10×10⁻³/5.0 = 2.0×10⁻³ A = 2.0 mA. (d) Away from 1000 kHz, X_L = ωL and X_C = 1/ωC no longer cancel: at higher frequencies X_L grows and X_C shrinks, at lower frequencies the reverse happens, so |X_L - X_C| becomes large. Since Z = √[R² + (X_L - X_C)²] then grows well above R, the current i_rms = e_rms/Z drops sharply, which is exactly how the circuit selects one station and suppresses the rest.
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