CLASS XII —
PHYSICS
Assignment: Alternating Current
(Chapter 7)
CBSE Board | Full
Chapter Coverage — Concepts + Numericals
Prepared by SD
PHYSICS
|
Name: ______________________ |
Time: 90 minutes |
Max Marks: 60 |
Useful constants/relations: XL = ωL, XC = 1/(ωC), Z = √(R² + (XL−XC)²), Irms = Iₘ/√2, Vrms = Vₘ/√2, ω₀ = 1/√(LC), Pavg = Vrms·Irms·cosφ
Section A — Very Short Answer Type (8 × 1 = 8 marks)
Q1. Why is AC preferred over DC for long-distance transmission of
electrical power?
[1]
Q2. What is the phase difference between voltage and current in a purely
resistive AC circuit? [1]
Q3. Define the term 'wattless current'. In which type of circuit does it
occur? [1]
Q4. Write the SI unit of inductive reactance and capacitive reactance. [1]
Q5. What is the value of power factor for a series LCR circuit at
resonance?
[1]
Q6. Why does a capacitor block DC but allow AC to pass through it? [1]
Q7. State the relation between the rms value and peak value of an
alternating current. [1]
Q8. Why can a choke coil (inductor) be used to control current in an AC
circuit without significant power loss, whereas a resistor cannot? [1]
Section B — Short Answer Type I (8 × 2 = 16 marks)
Q9. Draw the phasor diagram for a series LCR circuit in which XL >
XC, and label the phase angle φ. [2]
Q10. An alternating voltage V = Vm sin(ωt) is applied to a pure inductor
of inductance L. Derive an expression for the current flowing through it. [2]
Q11. Show mathematically that the average power consumed over
a full cycle in a purely capacitive AC circuit is zero. [2]
Q12. Define quality factor (Q-factor) of a series LCR
resonant circuit and write the expression for it in terms of L, C and R. [2]
Q13. Two students connect a bulb in series, first with a
capacitor and then with a resistor of equal DC resistance, to the same AC
source. Explain why the bulb glows with different brightness in the two cases. [2]
Q14. Explain, with reasoning, why a series LCR circuit is
said to behave as a pure resistor at resonance. [2]
Q15. An inductor and a resistor are connected in series to an
AC source. Explain why the current lags behind the voltage in this circuit, and
how the phase angle depends on the values of R and L. [2]
Q16. Distinguish between step-up and step-down transformers
on the basis of the turns ratio and the relationship between primary and
secondary voltage.
[2]
Section C — Short Answer Type II (6 × 3 = 18 marks)
Q17. A 100 Ω resistor is connected to a 220 V, 50 Hz AC
supply. [3]
(a) What
is the rms value of current in the circuit? (b) What is the net power consumed
over a full cycle? (c) Sketch the variation of voltage and current with time
over one cycle.
Q18. A pure inductor of inductance 25 mH is connected to a
220 V, 50 Hz source. Find (a) the inductive reactance, and (b) the rms current
in the circuit.
[3]
Q19. A 60 μF capacitor is connected to a 110 V, 60 Hz AC
supply. Determine (a) the capacitive reactance and (b) the rms current before
and after a dielectric is inserted between the plates of the capacitor
(reasoning only, no calculation needed for the second part). [3]
Q20. Explain, using energy considerations, how an LC circuit
sustains electrical oscillations, and why the oscillations are analogous to the
oscillations of a mechanical spring-mass system. [3]
Q21. A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is
connected to a variable-frequency 200 V AC source. Calculate the resonant
frequency of the circuit. [3]
Q22. Why does the reactance of an inductor increase with
frequency while that of a capacitor decreases with frequency? Justify using the
expressions for XL and XC and explain the practical significance for a choke
coil used at 50 Hz mains supply. [3]
Section D — Long Answer / Numerical Type (5 × 5 = 25 marks)
Q23. A series LCR circuit has R = 40 Ω, L = 0.2 H and C = 8 μF,
connected to a 220 V, 50 Hz AC source. [5]
(a)
Calculate XL and XC. (b) Calculate the impedance Z of the circuit. (c)
Calculate the current amplitude and the phase angle between current and
voltage. (d) State whether the circuit is predominantly inductive or capacitive
at this frequency.
Q24. Derive an expression for the impedance of a series LCR
circuit connected to an AC source using the phasor diagram method, and hence
obtain the expression for the current in the circuit. [5]
Q25. A series LCR circuit with L = 5.0 H, C = 80 μF and R =
40 Ω is
connected to a 230 V variable-frequency AC supply. [5]
(a) Find
the resonant frequency and the Q-factor of the circuit. (b) Obtain the peak
current at resonance. (c) Explain the significance of the Q-factor in relation
to the sharpness of resonance.
Q26. A transformer is used to step down 220 V AC mains supply
to 12 V for a device drawing a current of 4 A. If the transformer is 90%
efficient and the primary coil has 2000 turns, calculate: [5]
(a) the
number of turns in the secondary coil, (b) the current drawn from the mains,
and (c) the power input to the transformer.
Q27. Derive an expression for the average power dissipated in
a series LCR circuit over a complete cycle. Hence define the power factor of
the circuit, and identify the conditions under which the power factor is (i)
maximum and (ii) zero. [5]
Section E — Case Study Based Question (1 × 4 = 4 marks)
An
electrician wants to run a large inductive coil (motor winding) directly from
the 220 V, 50 Hz household mains. He observes that the coil gets warm and draws
a certain current, but if a suitable capacitor is connected in series with the
coil, the current drawn from the supply for the same power decreases and the
coil runs cooler for the same mechanical output.
Q28. (a) Explain why connecting a capacitor in series with
the inductive coil reduces the current drawn from the supply for the same real
power delivered.
[4]
(b) What
is this phenomenon called, and why is it important for power distribution
companies to encourage industries to maintain a power factor close to unity?
(c) If
the coil circuit alone has a lagging power factor of 0.6, and after adding the
capacitor the power factor becomes 0.9 (lagging), explain qualitatively what
has happened to the phase angle between voltage and current.
— CONSISTANCY IS
SHORTCUT TO SUCESS —
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