CHAPTER 7: ALTERNATING CURRENT
SAMPLE QUESTION PAPER (ASSIGNMENT-1)
- There are 33 questions in all. All questions are compulsory.
- Section A: 16 questions (12 MCQs and 4 Assertion-Reasoning) of 1 mark each.
- Section B: 5 short answer questions of 2 marks each.
- Section C: 7 short answer questions of 3 marks each.
- Section D: 2 case study-based questions of 4 marks each.
- Section E: 3 long answer questions of 5 marks each.
SECTION A (16 Marks)
Q.1. The SI unit of inductive reactance ($\displaystyle {X_L}$) is:
Explanation: Inductive reactance ($\displaystyle {X_L = \omega L}$) represents opposition offered by an inductor to AC flow and is measured in Ohms ($\Omega$).
Q.2. In a series LCR circuit, electrical resonance occurs when:
Explanation: Resonance occurs when inductive reactance equals capacitive reactance ($\displaystyle {X_L = X_C}$), making circuit impedance minimum ($\displaystyle {Z = R}$) and current maximum.
Q.3. In an ideal pure inductor connected across an AC voltage source, the current in the circuit:
Explanation: For an ideal inductor, $\displaystyle {V = V_0 \sin\omega t}$ produces current $\displaystyle {I = I_0 \sin(\omega t - \pi/2)}$, so current lags voltage by $\displaystyle {90^\circ}$.
Q.4. The average power consumed per cycle in a purely inductive or purely capacitive AC circuit is:
Explanation: Phase angle $\displaystyle {\phi = 90^\circ \implies \cos\phi = 0}$. Power dissipated $\displaystyle {P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos 90^\circ = 0}$.
Q.5. The RMS value of a sinusoidal AC voltage having a peak value of $\displaystyle {220\text{ V}}$ is:
Explanation: $\displaystyle {V_{\text{rms}} = {\frac{{V_0}}{{\sqrt{2}}}} = {\frac{{220}}{{1.414}}} \approx 155.6\text{ V}}$.
Q.6. In a series LCR circuit at resonance, the power factor is:
Explanation: At resonance $\displaystyle {X_L = X_C \implies Z = R}$. Power factor $\displaystyle {\cos\phi = R/Z = R/R = 1}$.
Q.7. The soft iron core of an electrical transformer is laminated in order to:
Explanation: Thin insulated laminations interrupt large circulating paths, increasing electrical resistance to eddy currents and reducing heating loss.
Q.8. A 'watt-less current' flows in an AC circuit when its power factor is:
Explanation: When current is $\displaystyle {90^\circ}$ out of phase with voltage ($\displaystyle {\phi = \pi/2}$), $\displaystyle {\cos\phi = 0}$ and average power consumed is zero.
Q.9. The EMF generated by an AC generator is $\displaystyle {V = V_m \sin\omega t}$. If angular frequency is doubled, peak EMF becomes:
Explanation: Peak EMF $\displaystyle {V_m = N B A \omega}$. Doubling $\omega$ doubles peak amplitude to $\displaystyle {2 V_m}$ and frequency to $\displaystyle {2\omega}$.
Q.10. Capacitive reactance ($\displaystyle {X_C}$) of a capacitor in an AC circuit varies with frequency $\displaystyle {f}$ as:
Explanation: $\displaystyle {X_C = {\frac{{1}}{{\omega C}}} = {\frac{{1}}{{2\pi f C}}} \implies X_C \propto {\frac{{1}}{{f}}}}$. Plot is a rectangular hyperbola.
Q.11. A choke coil is preferred over a rheostat to control alternating current in a circuit because:
Explanation: A choke coil has high inductance and negligible resistance ($\displaystyle {R \approx 0}$), so its power factor $\displaystyle {\cos\phi \approx 0}$, consuming almost zero power.
Q.12. In a series LCR circuit driven below the resonant frequency ($\displaystyle {f < f_0}$), the circuit behaves predominantly as:
Explanation: Below resonance ($\displaystyle {f < f_0}$), $\displaystyle {X_C = 1/(2\pi f C)}$ is greater than $\displaystyle {X_L = 2\pi f L}$, so current leads voltage.
Directions for Q.13 to Q.16:
(A) Both Assertion and Reason are true and Reason is correct explanation.
(B) Both Assertion and Reason are true but Reason is NOT correct explanation.
(C) Assertion is true but Reason is false.
(D) Both Assertion and Reason are false.
Q.13. Assertion (A): At resonance, the impedance of a series LCR circuit is purely resistive and minimum.
Reason (R): At resonance, inductive reactance cancels capacitive reactance ($\displaystyle {X_L = X_C}$).
Q.14. Assertion (A): In a purely capacitive circuit, current leads applied voltage by $\displaystyle {90^\circ}$.
Reason (R): The voltage across a capacitor cannot change instantaneously as charge accumulates.
Q.15. Assertion (A): At very high frequencies, a capacitor acts as a short circuit (conducting wire) for AC.
Reason (R): Capacitive reactance $\displaystyle {X_C = 1/(2\pi f C)}$ approaches zero as frequency $\displaystyle {f \to \infty}$.
Q.16. Assertion (A): A step-up transformer cannot be used as a step-down transformer.
Reason (R): A transformer works in only one direction.
Explanation: A transformer is a reversible passive device; interchanging primary and secondary connections transforms a step-up transformer into a step-down transformer.
SECTION B (10 Marks)
Q.17. When an AC source is connected across an ideal inductor, show on a graph the nature of variation of voltage and current over one complete cycle.
Q.18. A heating element is marked '$\displaystyle {210\text{ V}, 630\text{ W}}$'. Find its resistance when connected to a $\displaystyle {210\text{ V}}$ DC supply.
Q.19. Prove that an ideal inductor does not dissipate electrical power in an AC circuit.
Average power over full cycle: $\displaystyle {P_{\text{avg}} = -{\frac{{1}}{{2}}} V_0 I_0 \langle\sin(2\omega t)\rangle = 0}$ (since average of $\sin(2\omega t)$ over full cycle is 0).
Q.20. A $\displaystyle {15.0\,\mu\text{F}}$ capacitor is connected to a $\displaystyle {220\text{ V}, 50\text{ Hz}}$ AC source. Find capacitive reactance and RMS current.
RMS Current: $\displaystyle {I_{\text{rms}} = {\frac{{V_{\text{rms}}}{{X_C}}} = {\frac{{220}}{{212.2}}} \approx 1.04\text{ A}}$.
Q.21. Refer to the diagram showing two coaxial coils A and B.
(i) State the underlying physical principle.
(ii) Mention two factors on which induced current in coil B depends.
(i) Mutual Induction.
(ii) Factors: (1) Distance/spatial separation between coils, (2) Number of turns $\displaystyle {N_A, N_B}$ in both coils.
SECTION C (21 Marks)
Q.22. An electric lamp with negligible inductance is connected in series with a capacitor across an AC source. How does brightness change on reducing (i) capacitance $\displaystyle {C}$, (ii) source frequency $\displaystyle {f}$? Justify.
(i) Reducing $\displaystyle {C}$ increases $\displaystyle {X_C}$, which increases net impedance $\displaystyle {Z}$, reducing current $\displaystyle {I = V/Z}$. Brightness decreases.
(ii) Reducing frequency $\displaystyle {f}$ increases $\displaystyle {X_C}$, increasing $\displaystyle {Z}$ and reducing current. Brightness decreases.
Q.23. Using a phasor diagram, derive the expression for impedance $\displaystyle {Z}$ and phase angle $\displaystyle {\phi}$ in a series LCR AC circuit.
Net reactive voltage: $\displaystyle {V_L - V_C = I (X_L - X_C)}$ perpendicular to $\displaystyle {V_R}$.
Total voltage: $\displaystyle {V^2 = V_R^2 + (V_L - V_C)^2 = I^2 R^2 + I^2 (X_L - X_C)^2 = I^2 [R^2 + (X_L - X_C)^2]}$.
Impedance: $\displaystyle {Z = {\frac{{V}}{{I}}} = \sqrt{R^2 + (X_L - X_C)^2}}$.
Phase angle: $\displaystyle {\tan\phi = {\frac{{V_L - V_C}}{{V_R}}} = {\frac{{X_L - X_C}}{{R}}}}$.
Q.24. Calculate resonant frequency $\displaystyle {f_0}$ for a series LCR circuit with $\displaystyle {L = 0.1\text{ H}, C = 10\,\mu\text{F}, R = 10\,\Omega}$. What is impedance at resonance?
At resonance, $\displaystyle {X_L = X_C \implies Z = R = 10\,\Omega}$.
Q.25. A $\displaystyle {100\,\Omega}$ resistor is connected to a $\displaystyle {220\text{ V}, 50\text{ Hz}}$ AC supply. Calculate (i) RMS current, (ii) peak current, (iii) average power dissipated.
(i) $\displaystyle {I_{\text{rms}} = {\frac{{V_{\text{rms}}}{{R}}} = {\frac{{220}}{{100}}} = 2.2\text{ A}}$.
(ii) $\displaystyle {I_0 = I_{\text{rms}} \times \sqrt{2} = 2.2 \times 1.414 \approx 3.11\text{ A}}$.
(iii) $\displaystyle {P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} = 220 \times 2.2 = 484\text{ W}}$.
Q.26. On the basis of power dissipation, distinguish between resistance, reactance, and impedance in AC circuits.
1. Resistance ($\displaystyle {R}$): Causes continuous real power dissipation ($\displaystyle {P = I_{\text{rms}}^2 R}$) as heat.
2. Reactance ($\displaystyle {X}$): Opposition from pure inductors/capacitors; zero average power dissipation ($\displaystyle {P = 0}$) over a full cycle.
3. Impedance ($\displaystyle {Z}$): Combined opposition ($\displaystyle {Z = \sqrt{R^2 + X^2}}$); power dissipation depends on resistive component and power factor ($\displaystyle {P = V I \cos\phi}$).
Q.27. A circuit with an $\displaystyle {80\text{ mH}}$ inductor and $\displaystyle {60\,\mu\text{F}}$ capacitor in series is connected to a $\displaystyle {230\text{ V}, 50\text{ Hz}}$ AC supply with negligible resistance.
(a) Obtain RMS current.
(b) Obtain RMS potential drop across each element.
(c) What is average power transferred to inductor?
$\displaystyle {\omega = 2\pi f = 2\pi \times 50 = 100\pi \approx 314.16\text{ rad/s}}$.
$\displaystyle {X_L = \omega L = 314.16 \times 0.08 = 25.13\,\Omega}$.
$\displaystyle {X_C = {\frac{{1}}{{\omega C}}} = {\frac{{1}}{{314.16 \times (60 \times 10^{-6})}}} = 53.05\,\Omega}$.
Net reactance: $\displaystyle {|X_C - X_L| = |53.05 - 25.13| = 27.92\,\Omega}$.
(a) $\displaystyle {I_{\text{rms}} = {\frac{{V_{\text{rms}}}{{|X_C - X_L|}}} = {\frac{{230}}{{27.92}}} \approx 8.24\text{ A}}$.
(b) $\displaystyle {V_L = I_{\text{rms}} X_L = 8.24 \times 25.13 \approx 207.1\text{ V}}$; $\displaystyle {V_C = I_{\text{rms}} X_C = 8.24 \times 53.05 \approx 437.1\text{ V}}$.
(c) Average power transferred to ideal inductor is Zero ($\displaystyle {P = 0}$).
SECTION D: CASE STUDY QUESTIONS (8 Marks)
Q.29. Case Study 1: Series LCR Circuit and Resonance.
In a series LCR circuit connected to an AC source, impedance is $\displaystyle {Z = \sqrt{R^2 + (X_L - X_C)^2}}$. At resonant frequency $\displaystyle {f_0 = 1/(2\pi\sqrt{LC})}$, $\displaystyle {X_L = X_C}$, making impedance minimum ($\displaystyle {Z = R}$) and current maximum.
(i) Power factor at resonance is: (a) 0 (b) 1 (c) 0.5 (d) Infinite
(ii) If frequency $\displaystyle {\omega}$ increases beyond resonance, current: (a) Increases (b) Decreases (c) Remains constant
(iii) Resonance is sharpest for: (a) Low resistance (b) High resistance (c) Zero inductance
(iv) Quality factor $\displaystyle {Q}$ is given by: (a) $\displaystyle {\omega_0 L / R}$ (b) $\displaystyle {R / \omega_0 L}$ (c) $\displaystyle {\omega_0 R L}$
Q.30. Case Study 2: Transformer Construction & Transmission.
Transformers step up or step down AC voltage using mutual induction. In India, domestic supply is $\displaystyle {220\text{ V rms}}$ at $\displaystyle {50\text{ Hz}}$. Step-up transformers at power plants increase voltage to hundreds of kilovolts to minimize transmission line power loss ($\displaystyle {I^2 R}$).
(i) Turns ratio $\displaystyle {N_s/N_p}$ for stepping up from $\displaystyle {220\text{ V}}$ to $\displaystyle {11\text{ kV}}$ is: (a) 1:50 (b) 50:1 (c) 1:5
(ii) Why transformers do not operate on DC supply?
(iii) Role of soft iron core in transformers?
(iv) Primary current when secondary delivers $\displaystyle {10\text{ A}}$ at $\displaystyle {11\text{ kV}}$ (ideal, primary $\displaystyle {220\text{ V}}$)?
(i) (b) 50:1 ($\displaystyle {11000/220 = 50}$).
(ii) DC produces constant magnetic flux ($\displaystyle {d\Phi/dt = 0}$), so zero secondary EMF is induced.
(iii) Soft iron has high permeability to concentrate flux and low hysteresis loss.
(iv) Ideal power conservation: $\displaystyle {V_p I_p = V_s I_s \implies 220 \times I_p = 11000 \times 10 \implies I_p = 500\text{ A}}$.
SECTION E (15 Marks)
Q.31. (a) Derive expressions for impedance, current amplitude, and phase angle in a series LCR AC circuit using phasor method. Draw impedance triangle.
(b) Peak alternating voltage applied across $\displaystyle {50\,\Omega}$ resistor is $\displaystyle {10\text{ V}}$ at $\displaystyle {100\text{ Hz}}$. Find RMS current and equation for instantaneous current.
(a) Explain phase relationship between voltage and current in purely resistive, purely inductive, and purely capacitive AC circuits with phasor diagrams.
(b) Pure inductor of $\displaystyle {25\text{ mH}}$ is connected to $\displaystyle {220\text{ V}, 50\text{ Hz}}$ source. Find inductive reactance and RMS current.
(b) $\displaystyle {V_0 = 10\text{ V} \implies V_{\text{rms}} = 10/\sqrt{2} \approx 7.07\text{ V}}$.
$\displaystyle {I_{\text{rms}} = V_{\text{rms}}/R = 7.07/50 \approx 0.1414\text{ A} = 141.4\text{ mA}}$.
Peak current $\displaystyle {I_0 = V_0/R = 10/50 = 0.2\text{ A}}$. $\displaystyle {\omega = 2\pi (100) = 200\pi\text{ rad/s}}$.
Instantaneous current equation: $\displaystyle {i(t) = 0.2 \sin(200\pi t)\text{ A}}$.
OR Part (b): $\displaystyle {X_L = 2\pi f L = 2\pi \times 50 \times 0.025 = 2.5\pi \approx 7.85\,\Omega}$.
$\displaystyle {I_{\text{rms}} = V_{\text{rms}}/X_L = 220/7.85 \approx 28.03\text{ A}}$.
Q.32. (a) Draw graphs showing variation of inductive reactance ($\displaystyle {X_L}$) and capacitive reactance ($\displaystyle {X_C}$) with source frequency ($\displaystyle {f}$).
(b) Can RMS voltage across inductor or capacitor in series LCR circuit be greater than applied source voltage? Explain.
(a) Show that in an ideal transformer, stepping up voltage by a factor reduces current by the same factor.
(b) State four main causes of energy loss in practical transformers.
(b) Yes! Voltages $\displaystyle {V_L}$ and $\displaystyle {V_C}$ are $\displaystyle {180^\circ}$ out of phase, so they subtract vectorially ($\displaystyle {V = \sqrt{V_R^2 + (V_L - V_C)^2}}$). Individual reactive voltages can be higher than source voltage $\displaystyle {V}$.
OR Part:
(a) For ideal transformer, input power = output power $\displaystyle {V_p I_p = V_s I_s \implies {\frac{{V_s}}{{V_p}}} = {\frac{{I_p}}{{I_s}}} = {\frac{{N_s}}{{N_p}}}}$.
(b) Energy losses: (1) Copper loss ($\displaystyle {I^2 R}$ heating), (2) Eddy current loss in core, (3) Hysteresis loss, (4) Magnetic flux leakage.
Q.33. An input voltage $\displaystyle {V_{\text{in}} = 200 \sin(100\pi t)\text{ V}}$ is applied to an ideal step-down transformer having $\displaystyle {N_p = 1000}$ turns and $\displaystyle {N_s = 100}$ turns. The secondary load circuit has $\displaystyle {R = 4\,\Omega, X_C = 2\,\Omega, X_L = 6\,\Omega}$. Find:
(i) Peak output voltage across load,
(ii) Peak current in load circuit,
(iii) Average power supplied to load.
(i) Primary peak voltage $\displaystyle {V_{p0} = 200\text{ V}}$. Secondary peak voltage: $\displaystyle {V_{s0} = V_{p0} {\left( {\frac{{N_s}}{{N_p}}} \right)} = 200 \times {\left( {\frac{{100}}{{1000}}} \right)} = 20\text{ V}}$.
(ii) Load impedance: $\displaystyle {Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{4^2 + (6 - 2)^2} = \sqrt{16 + 16} = \sqrt{32} \approx 5.66\,\Omega}$.
Peak secondary current: $\displaystyle {I_{s0} = {\frac{{V_{s0}}}{{Z}}} = {\frac{{20}}{{\sqrt{32}}}} \approx 3.54\text{ A}}$.
(iii) Average power: $\displaystyle {P_{\text{avg}} = I_{\text{rms}}^2 R = {\left( {\frac{{I_{s0}}}{{\sqrt{2}}}} \right)}^2 R = {\frac{{I_{s0}^2 R}}{{2}}} = {\frac{{(3.54)^2 \times 4}}{{2}} \approx 25\text{ W}}}$.
ASSIGNMENT – 2 (MCQ PRACTICE)
1. An alternating current is given by $\displaystyle {I = 10 \sin(314 t)\text{ A}}$. Its peak value and frequency are:
Explanation: Peak $\displaystyle {I_0 = 10\text{ A}}$. Angular frequency $\displaystyle {\omega = 314 \implies f = 314 / (2\pi) \approx 50\text{ Hz}}$.
2. The ratio of inductive reactance to capacitive reactance in an AC circuit is:
Explanation: $\displaystyle {{\frac{{X_L}}{{X_C}}} = {\frac{{\omega L}}{{1/(\omega C)}}} = \omega^2 L C}$.
3. The average value of sinusoidal alternating current over one complete full cycle is:
Explanation: Positive and negative half-cycles cancel out exactly over a full time period.
4. In a circuit, current is $\displaystyle {I = 5 \sin\omega t\text{ A}}$ and voltage is $\displaystyle {V = 200 \cos\omega t\text{ V}}$. Phase difference between them is:
Explanation: $\displaystyle {V = 200 \sin(\omega t + \pi/2)}$. Thus voltage leads current by $\displaystyle {\pi/2}$ rad.
ASSIGNMENT – 3 (ASSERTION & REASON)
1. Assertion (A): A choke coil controls AC current with minimal power dissipation.
Reason (R): Power factor of an ideal choke coil is zero ($\displaystyle {\cos 90^\circ = 0}$).
2. Assertion (A): Quality factor $\displaystyle {Q}$ of a series LCR circuit increases as resistance $\displaystyle {R}$ decreases.
Reason (R): $\displaystyle {Q}$ is defined as $\displaystyle {\omega_0 L / R}$, so it is inversely proportional to resistance $\displaystyle {R}$.
ASSIGNMENT – 4 (CASE STUDY QUESTIONS)
Case Study: Household AC Circuit & Resistive Load
In a household, an electric heater of resistance $\displaystyle {50\,\Omega}$ is connected to a $\displaystyle {220\text{ V (RMS)}}$ supply at $\displaystyle {50\text{ Hz}}$.
1. RMS current through heater: (a) $\displaystyle {4.4\text{ A}}$ (b) $\displaystyle {2.2\text{ A}}$
2. Average power consumed: (a) $\displaystyle {968\text{ W}}$ (b) $\displaystyle {484\text{ W}}$
3. Peak voltage of supply ($\displaystyle {V_0}$): (a) $\displaystyle {220\text{ V}}$ (b) $\displaystyle {311\text{ V}}$
ASSIGNMENT – 5 (FORMULAE & CONCEPTUAL QUESTIONS)
CORE FORMULAE SUMMARY
Inductive: $\displaystyle {X_L = \omega L = 2\pi f L}$ | Capacitive: $\displaystyle {X_C = {\frac{{1}}{{\omega C}}} = {\frac{{1}}{{2\pi f C}}}}$
LCR Impedance: $\displaystyle {Z = \sqrt{R^2 + (X_L - X_C)^2}}$
Resonant Frequency: $\displaystyle {f_0 = {\frac{{1}}{{2\pi \sqrt{L C}}}}}$ | Quality Factor: $\displaystyle {Q = {\frac{{\omega_0 L}}{{R}}} = {\frac{{1}}{{R}}} \sqrt{{\frac{{L}}{{C}}}}}$
$\displaystyle {P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos\phi}$ | Power Factor: $\displaystyle {\cos\phi = {\frac{{R}}{{Z}}}}$
CONCEPTUAL SHORT QUESTIONS
1. Why is AC voltage preferred over DC voltage for long-distance electrical power transmission?
2. Why does a capacitor block direct current (DC) but allow alternating current (AC) to pass?
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