Chapter 7: Alternating Current Long Questions | physics Class 12 CBSE

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Chapter 7: Alternating Current Long Questions | physics Class 12 CBSE | ByteNotes.in

Long Answer

Medium difficulty • Structured explanation

Medium

Question 1

Long Form

Derive an expression for the current in a pure resistor connected to an ac source. Show that voltage and current are in phase, and obtain the average power dissipated.

Applying Kirchhoff's loop rule to a resistor R connected to v = vm sin(ωt): vm sin(ωt) = iR, giving i = (vm/R) sin(ωt) = im sin(ωt), where im = vm/R. This is Ohm's law valid for both ac and dc.
Since both v and i are proportional to sin(ωt) with no phase shift, voltage and current reach their maximum, minimum, and zero values simultaneously — they are in phase (phase difference = 0).
Instantaneous power: p = i²R = im²R sin²(ωt). Average over a cycle: P = im²R × (1/2) = (1/2)im²R, since the average of sin²(ωt) = 1/2.
Defining rms current I = im/√2, average power becomes P = I²R, identical in form to dc power, justifying the concept of rms values.
Thus, a pure resistor dissipates real power in an ac circuit, unlike pure inductors or capacitors, and the rms current is the equivalent dc current producing the same heating effect.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 2

Long Form

Derive the expression for current in a pure inductor connected to an ac source. Compare the phase of current with voltage, define inductive reactance, and show that average power dissipated is zero.

For a pure inductor L with v = vm sin(ωt), Kirchhoff's loop rule gives: v − L(di/dt) = 0, so di/dt = (vm/L) sin(ωt). Integrating: i = −(vm/ωL) cos(ωt) = (vm/ωL) sin(ωt − π/2) = im sin(ωt − π/2).
The current lags the voltage by π/2 (90°). Comparing with dc case, ωL plays the role of resistance; it is called inductive reactance XL = ωL (in ohms), and im = vm/XL.
XL = ωL is directly proportional to both angular frequency ω and inductance L. Higher frequency or larger inductance means greater opposition to ac current.
Instantaneous power: pL = iv = im sin(ωt − π/2) × vm sin(ωt) = −(imvm/2) sin(2ωt). Average of sin(2ωt) over a complete cycle is zero.
Therefore, average power PL = 0. Energy stored in the magnetic field during one half cycle is completely returned in the next, so no net energy is dissipated by a pure inductor.

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Long Answer

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Hard

Question 3

Long Form

Using the phasor diagram method, derive the expression for impedance and phase angle of a series LCR circuit. State the conditions under which the circuit is inductive or capacitive.

In a series LCR circuit with current i = im sin(ωt + φ), the voltage amplitudes are: vRm = imR (in phase with I), vLm = imXL (leads I by π/2), vCm = imXC (lags I by π/2).
Since VL and VC are antiparallel, their net phasor has magnitude |vCm − vLm|. The total voltage phasor V is the hypotenuse of a right triangle with sides VR and (VL − VC): vm² = vRm² + (vCm − vLm)².
Substituting: vm = im√[R² + (XC − XL)²]. By analogy with Ohm's law, impedance Z = √[R² + (XC − XL)²] and im = vm/Z.
The phase angle: tan φ = (vCm − vLm)/vRm = (XC − XL)/R. If XC > XL, φ > 0 (current leads voltage, circuit is capacitive). If XC < XL, φ < 0 (current lags voltage, circuit is inductive).
The impedance diagram is a right triangle with Z as hypotenuse, R as base, and (XC − XL) as perpendicular, providing a geometric representation of circuit behavior.

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Long Answer

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Hard

Question 4

Long Form

Explain the phenomenon of resonance in a series LCR circuit. Derive the resonant frequency, describe the variation of current amplitude with frequency, and discuss how this principle is applied in radio tuning.

In a series LCR circuit, current amplitude im = vm/√[R² + (XC − XL)²] is maximum when the denominator is minimum. This occurs when XC = XL, i.e., 1/(ω₀C) = ω₀L, giving resonant frequency ω₀ = 1/√(LC).
At resonance, Z = R (minimum), im = vm/R (maximum), and φ = 0 (current and voltage are in phase). The voltages across L and C are equal and opposite, canceling each other completely.
As ω is varied, im increases toward the resonant peak and falls on either side. Lower resistance R gives a sharper, taller resonance peak, while higher R gives a flatter, broader peak (as shown by comparing R = 100 Ω and R = 200 Ω cases).
Resonance can only occur when both L and C are present in the circuit. In RL or RC circuits, there is no cancellation of reactive voltages, so resonance is impossible.
In radio tuning, the antenna feeds signals of many frequencies to the tuning circuit (L and variable C). By adjusting C, ω₀ is made equal to the desired station's frequency, maximizing current for that signal while suppressing others.

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Long Answer

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Hard

Question 5

Long Form

Discuss power in ac circuits for the following cases: (i) pure resistor, (ii) pure inductor or capacitor, (iii) series LCR circuit at resonance. Define power factor and explain how it can be improved.

For a pure resistor: φ = 0, cos φ = 1, P = VI. All power supplied is dissipated as heat. This is the maximum power case.
For a pure inductor or capacitor: φ = π/2, cos φ = 0, P = 0. Current flows but no power is dissipated — this is called wattless current. Energy alternates between source and the reactive element.
For a series LCR circuit: P = VI cos φ, where φ = tan⁻¹[(XC − XL)/R]. Power is dissipated only in the resistance R; inductors and capacitors contribute no net power dissipation.
At resonance in LCR: XC = XL, φ = 0, cos φ = 1, P = I²R — maximum power is dissipated. The power factor equals 1.
Power factor (cos φ = R/Z) can be improved toward 1 by connecting a capacitor in parallel to neutralize the lagging wattless component of current (Iq), reducing Z toward R and improving efficiency of power delivery.

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Long Answer

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Medium

Question 6

Long Form

Describe the construction and working of an ideal transformer. Derive the voltage and current transformation ratios and distinguish between step-up and step-down transformers.

A transformer consists of a primary coil (Np turns) and a secondary coil (Ns turns) wound on a soft-iron core. An alternating voltage applied to the primary creates a changing magnetic flux that links both coils.
For ideal transformer: induced emf in secondary εs = −Ns(dφ/dt) and back emf in primary εp = −Np(dφ/dt). Since εp = vp and εs = vs, dividing: vs/vp = Ns/Np (voltage transformation ratio).
For 100% efficiency, input power = output power: ipvp = isvs. Combined with the voltage ratio: ip/is = vs/vp = Ns/Np, giving the current transformation ratio.
Step-up transformer: Ns > Np, so Vs > Vp and Is < Ip — voltage increases, current decreases. Step-down transformer: Ns < Np, so Vs < Vp and Is > Ip — voltage decreases, current increases.
The three key assumptions for deriving these ratios are: (i) primary resistance and current are small; (ii) the same flux links both coils (negligible flux leakage); (iii) secondary current is small (open or lightly loaded circuit).

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Long Answer

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Medium

Question 7

Long Form

Explain why ac voltage is preferred over dc voltage for long-distance power transmission. Describe the role of transformers at different stages from generation to consumer.

AC is preferred because transformers can efficiently step up or step down ac voltages using mutual induction; this is not possible for dc without complex and lossy conversion equipment.
Power loss in transmission lines is P_loss = I²R. For a given power P = IV, high voltage means low current, drastically reducing I²R losses over long resistance wires.
At the generating station, a step-up transformer raises voltage (e.g., to hundreds of kV), reducing current and minimizing transmission losses over long distances.
At an area substation near consumers, a step-down transformer reduces voltage to intermediate levels (e.g., 11 kV). Further step-down transformers at distribution substations and utility poles reduce it to 240 V for home use.
A well-designed transformer has efficiency exceeding 95%, making the overall transmission system economically viable. Energy losses in transformers (eddy currents, hysteresis, winding resistance, flux leakage) are minimized by careful design.

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Long Answer

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Medium

Question 8

Long Form

Derive the expression for current in a pure capacitor connected to an ac source. Explain how capacitive reactance differs from inductive reactance in its frequency dependence, and show average power is zero.

For a pure capacitor C with v = vm sin(ωt): from Kirchhoff's loop rule, q/C = vm sin(ωt), so q = Cvm sin(ωt). Current i = dq/dt = ωCvm cos(ωt) = ωCvm sin(ωt + π/2) = im sin(ωt + π/2).
Current leads voltage by π/2. The amplitude im = ωCvm = vm/XC, where capacitive reactance XC = 1/(ωC), measured in ohms.
Unlike inductive reactance XL = ωL (which increases with frequency), capacitive reactance XC = 1/(ωC) decreases as frequency increases. Higher frequency means lower opposition and more current through a capacitor.
Instantaneous power: pC = iv = im cos(ωt) × vm sin(ωt) = (imvm/2) sin(2ωt). Average of sin(2ωt) over a complete cycle is zero.
Therefore, average power PC = 0. Energy stored in the electric field of the capacitor during charging is completely returned during discharging — no net energy dissipation occurs in a pure capacitor.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 9

Long Form

Compare the behavior of ac circuits containing (i) only R, (ii) only L, and (iii) only C with respect to phase relationship, reactance/resistance, and average power dissipated.

Pure R circuit: current and voltage are in phase (φ = 0). Opposition is resistance R (independent of frequency). Average power P = I²R ≠ 0; maximum power is dissipated.
Pure L circuit: current lags voltage by π/2 (φ = −π/2). Opposition is inductive reactance XL = ωL, which increases with frequency. Average power PL = 0; energy is stored and returned from the magnetic field.
Pure C circuit: current leads voltage by π/2 (φ = +π/2). Opposition is capacitive reactance XC = 1/(ωC), which decreases with frequency. Average power PC = 0; energy is stored and returned from the electric field.
In both L and C circuits, the current is termed 'wattless current' as it flows without net power dissipation, while in R, the current does real work (Joule heating).
This comparison shows that only resistance dissipates electrical energy in ac circuits; pure inductors and capacitors are reactive elements that store and return energy cyclically.

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Long Answer

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Hard

Question 10

Long Form

Explain the concept of impedance in a series LCR circuit. Using an impedance diagram, discuss how the circuit's nature changes with the relative values of XL and XC.

Impedance Z = √[R² + (XC − XL)²] is the total effective opposition to ac current in a series LCR circuit. It combines resistance and net reactance, measured in ohms, with im = vm/Z.
The impedance diagram is a right triangle: R along the base, (XC − XL) along the perpendicular, and Z as the hypotenuse. The angle φ between Z and R gives the phase angle: tan φ = (XC − XL)/R.
When XC > XL: (XC − XL) > 0, φ > 0, the circuit is predominantly capacitive, and current leads voltage.
When XC < XL: (XC − XL) < 0, φ < 0, the circuit is predominantly inductive, and current lags voltage.
When XC = XL (resonance): (XC − XL) = 0, Z = R (minimum), φ = 0, power factor = 1, and current is maximum. The impedance diagram collapses to just the resistance R.

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Chapter 7: Alternating Current | Long Questions

Derive an expression for the current in a pure resistor connected to an ac source. Show that voltage and current are in phase, and obtain the average power dissipated.

Derive the expression for current in a pure inductor connected to an ac source. Compare the phase of current with voltage, define inductive reactance, and show that average power dissipated is zero.

Using the phasor diagram method, derive the expression for impedance and phase angle of a series LCR circuit. State the conditions under which the circuit is inductive or capacitive.

Explain the phenomenon of resonance in a series LCR circuit. Derive the resonant frequency, describe the variation of current amplitude with frequency, and discuss how this principle is applied in radio tuning.

Discuss power in ac circuits for the following cases: (i) pure resistor, (ii) pure inductor or capacitor, (iii) series LCR circuit at resonance. Define power factor and explain how it can be improved.

Describe the construction and working of an ideal transformer. Derive the voltage and current transformation ratios and distinguish between step-up and step-down transformers.

Explain why ac voltage is preferred over dc voltage for long-distance power transmission. Describe the role of transformers at different stages from generation to consumer.

Derive the expression for current in a pure capacitor connected to an ac source. Explain how capacitive reactance differs from inductive reactance in its frequency dependence, and show average power is zero.

Compare the behavior of ac circuits containing (i) only R, (ii) only L, and (iii) only C with respect to phase relationship, reactance/resistance, and average power dissipated.

Explain the concept of impedance in a series LCR circuit. Using an impedance diagram, discuss how the circuit's nature changes with the relative values of XL and XC.

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