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Alternating Current NCERT Solutions & Important Questions
Welcome! If Chapter 7 Alternating Current is giving you tension, don't worry. This guide covers the Updated NCERT Solutions for Class 12 Physics Chapter 7, simplifying phasors, LCR circuits, and transformers. Perfect for your CBSE Board Exam 2026 and competitive prep like JEE/NEET. We will master these concepts logically—no blind rote memorization required. Toh chaliye, padhai shuru karein!
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Chapter NameAlternating Current (Chapter 7)
SubjectPhysics
ClassClass 12
BoardCBSE (2026-27 Syllabus)
DifficultyModerate to High (Formula & Concept heavy)
Exam Weightage~7-8 Marks (Combined with EMI)
Learning Objectives
After completing this chapter, students will be able to:
- Understand the difference between alternating current (AC) and direct current (DC).
- Calculate root mean square (rms) and peak values of current and voltage.
- Analyze AC circuits containing resistors, inductors, and capacitors individually and in series (LCR).
- Determine the conditions for resonance in an LCR circuit and calculate the resonant frequency.
- Understand power factor and the concept of wattless current.
- Explain the working principle, construction, and energy losses of a transformer.
Key Concepts & Formulas
- Alternating Voltage/Current: Changes continuously in magnitude and periodically in direction.
$$V = V_m \sin(\omega t)$$
$$I = I_m \sin(\omega t)$$ - RMS Values: The effective value of AC.
$$I_{rms} = \frac{I_m}{\sqrt{2}} = 0.707 I_m$$ - Reactance ($X$): Opposition offered by an inductor or capacitor to AC.
Inductive Reactance: $$X_L = \omega L = 2\pi fL$$
Capacitive Reactance: $$X_C = \frac{1}{\omega C} = \frac{1}{2\pi fC}$$ - Impedance ($Z$): Total opposition in an LCR circuit.
$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$ - Resonant Frequency ($f_r$): Occurs when $X_L = X_C$.
$$f_r = \frac{1}{2\pi \sqrt{LC}}$$ - Power in AC Circuit:
$$P_{avg} = V_{rms} I_{rms} \cos\phi$$
(Where $\cos\phi = \frac{R}{Z}$ is the power factor).
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Full NCERT Solutions
Question 7.1: A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle?
Answer:
Step 1: Note down given values.
$R = 100~\Omega$, $V_{rms} = 220\text{ V}$, $f = 50\text{ Hz}$.
(a) Calculating the rms current using Ohm's law:
$$I_{rms} = \frac{V_{rms}}{R} = \frac{220}{100} = 2.2\text{ A}$$
(b) Calculating net power consumed in a purely resistive circuit:
$$P = V_{rms} \times I_{rms} = 220 \times 2.2 = 484\text{ W}$$
Step 1: Note down given values.
$R = 100~\Omega$, $V_{rms} = 220\text{ V}$, $f = 50\text{ Hz}$.
(a) Calculating the rms current using Ohm's law:
$$I_{rms} = \frac{V_{rms}}{R} = \frac{220}{100} = 2.2\text{ A}$$
(b) Calculating net power consumed in a purely resistive circuit:
$$P = V_{rms} \times I_{rms} = 220 \times 2.2 = 484\text{ W}$$
Question 7.2: (a) The peak voltage of an ac supply is 300 V. What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A. What is the peak current?
Answer:
Step 1: Calculate rms voltage (a).
Given: Peak voltage $V_m = 300\text{ V}$
$$V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{300}{1.414} = 212.1\text{ V}$$
Step 2: Calculate peak current (b).
Given: $I_{rms} = 10\text{ A}$
$$I_m = I_{rms} \times \sqrt{2} = 10 \times 1.414 = 14.14\text{ A}$$
Step 1: Calculate rms voltage (a).
Given: Peak voltage $V_m = 300\text{ V}$
$$V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{300}{1.414} = 212.1\text{ V}$$
Step 2: Calculate peak current (b).
Given: $I_{rms} = 10\text{ A}$
$$I_m = I_{rms} \times \sqrt{2} = 10 \times 1.414 = 14.14\text{ A}$$
Question 7.3: A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Answer:
Step 1: Note the given data.
$L = 44\text{ mH} = 44 \times 10^{-3}\text{ H}$, $V_{rms} = 220\text{ V}$, $f = 50\text{ Hz}$.
Step 2: Calculate Inductive Reactance ($X_L$).
$$X_L = 2\pi fL = 2 \times 3.14 \times 50 \times 44 \times 10^{-3} = 13.82~\Omega$$
Step 3: Calculate rms current.
$$I_{rms} = \frac{V_{rms}}{X_L} = \frac{220}{13.82} = 15.9\text{ A}$$
Step 1: Note the given data.
$L = 44\text{ mH} = 44 \times 10^{-3}\text{ H}$, $V_{rms} = 220\text{ V}$, $f = 50\text{ Hz}$.
Step 2: Calculate Inductive Reactance ($X_L$).
$$X_L = 2\pi fL = 2 \times 3.14 \times 50 \times 44 \times 10^{-3} = 13.82~\Omega$$
Step 3: Calculate rms current.
$$I_{rms} = \frac{V_{rms}}{X_L} = \frac{220}{13.82} = 15.9\text{ A}$$
Question 7.4: A 60 µF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Answer:
Step 1: Note the given data.
$C = 60~\mu\text{F} = 60 \times 10^{-6}\text{ F}$, $V_{rms} = 110\text{ V}$, $f = 60\text{ Hz}$.
Step 2: Calculate Capacitive Reactance ($X_C$).
$$X_C = \frac{1}{2\pi fC} = \frac{1}{2 \times 3.14 \times 60 \times 60 \times 10^{-6}} = 44.2~\Omega$$
Step 3: Calculate rms current.
$$I_{rms} = \frac{V_{rms}}{X_C} = \frac{110}{44.2} = 2.49\text{ A}$$
Step 1: Note the given data.
$C = 60~\mu\text{F} = 60 \times 10^{-6}\text{ F}$, $V_{rms} = 110\text{ V}$, $f = 60\text{ Hz}$.
Step 2: Calculate Capacitive Reactance ($X_C$).
$$X_C = \frac{1}{2\pi fC} = \frac{1}{2 \times 3.14 \times 60 \times 60 \times 10^{-6}} = 44.2~\Omega$$
Step 3: Calculate rms current.
$$I_{rms} = \frac{V_{rms}}{X_C} = \frac{110}{44.2} = 2.49\text{ A}$$
Question 7.5: In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle? Explain your answer.
Answer:
The net power absorbed in both circuits is zero.
Explanation:
* In a purely inductive circuit (7.3), current lags the voltage by $90^{\circ}$. Power factor $\cos(90^{\circ}) = 0$.
* In a purely capacitive circuit (7.4), current leads the voltage by $90^{\circ}$. Power factor $\cos(90^{\circ}) = 0$. Hence, $P = V_{rms} I_{rms} \cos\phi = 0$.
The net power absorbed in both circuits is zero.
Explanation:
* In a purely inductive circuit (7.3), current lags the voltage by $90^{\circ}$. Power factor $\cos(90^{\circ}) = 0$.
* In a purely capacitive circuit (7.4), current leads the voltage by $90^{\circ}$. Power factor $\cos(90^{\circ}) = 0$. Hence, $P = V_{rms} I_{rms} \cos\phi = 0$.
Question 7.6: Obtain the resonant frequency $\omega_r$ of a series LCR circuit with $L = 2.0\text{ H}$, $C = 32~\mu\text{F}$ and $R = 10~\Omega$. What is the Q-value of this circuit?
Answer:
Step 1: Calculate Resonant angular frequency ($\omega_r$).
Given: $L = 2.0\text{ H}$, $C = 32 \times 10^{-6}\text{ F}$, $R = 10~\Omega$.
$$\omega_r = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{2 \times 32 \times 10^{-6}}} = \frac{1}{\sqrt{64 \times 10^{-6}}} = \frac{1}{8 \times 10^{-3}} = 125\text{ rad/s}$$
Step 2: Calculate Quality factor (Q-value).
$$Q = \frac{1}{R} \sqrt{\frac{L}{C}} = \frac{1}{10} \sqrt{\frac{2.0}{32 \times 10^{-6}}} = \frac{1}{10} \sqrt{62500} = \frac{250}{10} = 25$$
Step 1: Calculate Resonant angular frequency ($\omega_r$).
Given: $L = 2.0\text{ H}$, $C = 32 \times 10^{-6}\text{ F}$, $R = 10~\Omega$.
$$\omega_r = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{2 \times 32 \times 10^{-6}}} = \frac{1}{\sqrt{64 \times 10^{-6}}} = \frac{1}{8 \times 10^{-3}} = 125\text{ rad/s}$$
Step 2: Calculate Quality factor (Q-value).
$$Q = \frac{1}{R} \sqrt{\frac{L}{C}} = \frac{1}{10} \sqrt{\frac{2.0}{32 \times 10^{-6}}} = \frac{1}{10} \sqrt{62500} = \frac{250}{10} = 25$$
Question 7.7: A charged 30 µF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Answer:
Step 1: Apply the formula for LC oscillations.
Given: $C = 30 \times 10^{-6}\text{ F}$, $L = 27 \times 10^{-3}\text{ H}$.
$$\omega = \frac{1}{\sqrt{LC}} \text{ rad/s}$$
$$\omega = \frac{1}{\sqrt{27 \times 10^{-3} \times 30 \times 10^{-6}}} = \frac{1}{\sqrt{81 \times 10^{-8}}}$$
$$\omega = \frac{1}{9 \times 10^{-4}} = 1.11 \times 10^3\text{ rad/s}$$
Step 1: Apply the formula for LC oscillations.
Given: $C = 30 \times 10^{-6}\text{ F}$, $L = 27 \times 10^{-3}\text{ H}$.
$$\omega = \frac{1}{\sqrt{LC}} \text{ rad/s}$$
$$\omega = \frac{1}{\sqrt{27 \times 10^{-3} \times 30 \times 10^{-6}}} = \frac{1}{\sqrt{81 \times 10^{-8}}}$$
$$\omega = \frac{1}{9 \times 10^{-4}} = 1.11 \times 10^3\text{ rad/s}$$
Question 7.8: Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?
Answer:
Step 1: Calculate Initial Energy.
Given: Initial charge $Q_0 = 6\text{ mC} = 6 \times 10^{-3}\text{ C}$.
Initial energy is entirely electrostatic (stored in the capacitor):
$$E = \frac{Q_0^2}{2C} = \frac{(6 \times 10^{-3})^2}{2 \times 30 \times 10^{-6}} = \frac{36 \times 10^{-6}}{60 \times 10^{-6}} = 0.6\text{ J}$$
Step 2: Determine total energy at a later time.
According to the law of conservation of energy, the total energy of the LC circuit remains constant (0.6 J) at any later time, simply transferring between the capacitor and inductor.
Step 1: Calculate Initial Energy.
Given: Initial charge $Q_0 = 6\text{ mC} = 6 \times 10^{-3}\text{ C}$.
Initial energy is entirely electrostatic (stored in the capacitor):
$$E = \frac{Q_0^2}{2C} = \frac{(6 \times 10^{-3})^2}{2 \times 30 \times 10^{-6}} = \frac{36 \times 10^{-6}}{60 \times 10^{-6}} = 0.6\text{ J}$$
Step 2: Determine total energy at a later time.
According to the law of conservation of energy, the total energy of the LC circuit remains constant (0.6 J) at any later time, simply transferring between the capacitor and inductor.
Question 7.9: A series LCR circuit with $R = 20~\Omega$, $L = 1.5\text{ H}$ and $C = 35~\mu\text{F}$ is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?
Answer:
Step 1: Calculate current at resonance.
At the natural (resonant) frequency, $X_L = X_C$, so the impedance is purely resistive ($Z = R$).
Given: $R = 20~\Omega$, $V_{rms} = 200\text{ V}$.
$$I_{rms} = \frac{V_{rms}}{R} = \frac{200}{20} = 10\text{ A}$$
Step 2: Calculate Average power.
$$P = I_{rms}^2 R = (10)^2 \times 20 = 100 \times 20 = 2000\text{ W} = 2\text{ kW}$$
Step 1: Calculate current at resonance.
At the natural (resonant) frequency, $X_L = X_C$, so the impedance is purely resistive ($Z = R$).
Given: $R = 20~\Omega$, $V_{rms} = 200\text{ V}$.
$$I_{rms} = \frac{V_{rms}}{R} = \frac{200}{20} = 10\text{ A}$$
Step 2: Calculate Average power.
$$P = I_{rms}^2 R = (10)^2 \times 20 = 100 \times 20 = 2000\text{ W} = 2\text{ kW}$$
Question 7.10: A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor?
Answer:
Step 1: State the formula.
Given: $L = 200~\mu\text{H} = 200 \times 10^{-6}\text{ H}$. $f_1 = 800\text{ kHz} = 800 \times 10^3\text{ Hz}$. $f_2 = 1200\text{ kHz} = 1200 \times 10^3\text{ Hz}$.
Using $f = \frac{1}{2\pi \sqrt{LC}}$, we get $C = \frac{1}{4\pi^2 f^2 L}$
Step 2: Calculate for $f_1 = 800\text{ kHz}$.
$$C_1 = \frac{1}{4 \times (3.14)^2 \times (800 \times 10^3)^2 \times 200 \times 10^{-6}} \approx 198\text{ pF}$$
Step 3: Calculate for $f_2 = 1200\text{ kHz}$.
$$C_2 = \frac{1}{4 \times (3.14)^2 \times (1200 \times 10^3)^2 \times 200 \times 10^{-6}} \approx 88\text{ pF}$$
Final Answer: Range of capacitor is 88 pF to 198 pF.
Step 1: State the formula.
Given: $L = 200~\mu\text{H} = 200 \times 10^{-6}\text{ H}$. $f_1 = 800\text{ kHz} = 800 \times 10^3\text{ Hz}$. $f_2 = 1200\text{ kHz} = 1200 \times 10^3\text{ Hz}$.
Using $f = \frac{1}{2\pi \sqrt{LC}}$, we get $C = \frac{1}{4\pi^2 f^2 L}$
Step 2: Calculate for $f_1 = 800\text{ kHz}$.
$$C_1 = \frac{1}{4 \times (3.14)^2 \times (800 \times 10^3)^2 \times 200 \times 10^{-6}} \approx 198\text{ pF}$$
Step 3: Calculate for $f_2 = 1200\text{ kHz}$.
$$C_2 = \frac{1}{4 \times (3.14)^2 \times (1200 \times 10^3)^2 \times 200 \times 10^{-6}} \approx 88\text{ pF}$$
Final Answer: Range of capacitor is 88 pF to 198 pF.
Question 7.11: Figure shows a series LCR circuit connected to a variable frequency 230 V source. $L = 5.0\text{ H}$, $C = 80~\mu\text{F}$, $R = 40~\Omega$. (a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
Answer:
Step 1: Calculate resonant frequency (a).
$$\omega_r = \frac{1}{\sqrt{LC}}$$
$$\omega_r = \frac{1}{\sqrt{5 \times 80 \times 10^{-6}}} = \frac{1}{\sqrt{400 \times 10^{-6}}} = \frac{1}{0.02} = 50\text{ rad/s}$$
Frequency $f_r = \frac{\omega_r}{2\pi} = \frac{50}{2 \times 3.14} = 7.96\text{ Hz}$.
Step 2: Calculate impedance and amplitude of current (b).
At resonance, Impedance $Z = R = 40~\Omega$.
Amplitude of current (Peak current $I_m$):
$$I_{rms} = \frac{V_{rms}}{Z} = \frac{230}{40} = 5.75\text{ A}$$
$$I_m = I_{rms} \times \sqrt{2} = 5.75 \times 1.414 = 8.13\text{ A}$$
Step 1: Calculate resonant frequency (a).
$$\omega_r = \frac{1}{\sqrt{LC}}$$
$$\omega_r = \frac{1}{\sqrt{5 \times 80 \times 10^{-6}}} = \frac{1}{\sqrt{400 \times 10^{-6}}} = \frac{1}{0.02} = 50\text{ rad/s}$$
Frequency $f_r = \frac{\omega_r}{2\pi} = \frac{50}{2 \times 3.14} = 7.96\text{ Hz}$.
Step 2: Calculate impedance and amplitude of current (b).
At resonance, Impedance $Z = R = 40~\Omega$.
Amplitude of current (Peak current $I_m$):
$$I_{rms} = \frac{V_{rms}}{Z} = \frac{230}{40} = 5.75\text{ A}$$
$$I_m = I_{rms} \times \sqrt{2} = 5.75 \times 1.414 = 8.13\text{ A}$$
Extra Important Questions (Board Style 2026)
Multiple Choice Questions (MCQs)
1. In a purely inductive AC circuit, the phase difference between voltage and current is:
A) 0
B) $\pi/4$
C) $\pi/2$
D) $\pi$
Answer: C (Current lags voltage by 90° or $\pi/2$).
2. The core of a transformer is laminated to reduce:
A) Copper loss
B) Magnetic flux leak
C) Hysteresis loss
D) Eddy current loss
Answer: D
3. At resonance, the power factor of a series LCR circuit is:
A) 0
B) 0.5
C) 1
D) Infinity
Answer: C (Because phase angle $\phi=0$ so $\cos(0)=1$).
Short Answer Questions
4. Why is DC not used in transformers?
Answer: Transformers work on the principle of mutual induction, which requires a changing magnetic flux. DC provides a steady current, meaning no changing flux, so no EMF is induced in the secondary coil.
5. Define Wattless Current.
Answer: When the power dissipated in an AC circuit is zero despite the flow of current (like in purely inductive or capacitive circuits where phase angle is 90°), the current is called wattless current.
6. What is the significance of the Q-factor in an LCR circuit?
Answer: The Q-factor (Quality factor) measures the sharpness of resonance. A higher Q-factor means a sharper resonance curve, which provides better selectivity/tuning in radio receivers.
7. A light bulb and an open coil inductor are connected to an AC source through a key. What happens to the brightness of the bulb when an iron rod is inserted into the inductor?
Answer: The brightness decreases. Inserting the iron rod increases the magnetic field, which increases the inductance ($L$). This increases inductive reactance ($X_L$), increasing total impedance ($Z$). Hence, current decreases.
Long Answer Questions
8. Derive the expression for impedance in a series LCR circuit using a phasor diagram.
Answer:
Step 1: Let an AC voltage $V = V_m \sin\omega t$ be applied to L, C, and R in series.
Step 2: Voltage across resistor ($V_R$) is in phase with current I.
Step 3: Voltage across inductor ($V_L$) leads I by 90°.
Step 4: Voltage across capacitor ($V_C$) lags I by 90°.
Step 5: Using a phasor diagram, $V_L$ and $V_C$ are opposite. Net reactive voltage is ($V_L - V_C$).
Step 6: Resultant voltage $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.
Step 7: Since $V = IZ$, $V_R = IR$, $V_L = IX_L$, $V_C = IX_C$.
Substituting, we get $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Step 1: Let an AC voltage $V = V_m \sin\omega t$ be applied to L, C, and R in series.
Step 2: Voltage across resistor ($V_R$) is in phase with current I.
Step 3: Voltage across inductor ($V_L$) leads I by 90°.
Step 4: Voltage across capacitor ($V_C$) lags I by 90°.
Step 5: Using a phasor diagram, $V_L$ and $V_C$ are opposite. Net reactive voltage is ($V_L - V_C$).
Step 6: Resultant voltage $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.
Step 7: Since $V = IZ$, $V_R = IR$, $V_L = IX_L$, $V_C = IX_C$.
Substituting, we get $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
9. Explain the principle, construction, and working of a Transformer. Mention its energy losses.
Answer:
* Principle: Mutual Induction.
* Construction: Primary and secondary coils wound over a laminated soft iron core.
* Working: AC in the primary creates a changing magnetic flux, which links with the secondary coil and induces an EMF. $\frac{V_s}{V_p} = \frac{N_s}{N_p}$.
* Energy Losses: Flux leakage, resistance of windings (copper loss), eddy currents (iron loss), and hysteresis.
* Principle: Mutual Induction.
* Construction: Primary and secondary coils wound over a laminated soft iron core.
* Working: AC in the primary creates a changing magnetic flux, which links with the secondary coil and induces an EMF. $\frac{V_s}{V_p} = \frac{N_s}{N_p}$.
* Energy Losses: Flux leakage, resistance of windings (copper loss), eddy currents (iron loss), and hysteresis.
10. What is resonance in an AC circuit? Derive the formula for resonant frequency and explain its application.
Answer:
Step 1: Definition.
Resonance occurs when the current amplitude in an LCR circuit becomes maximum, which happens when $X_L = X_C$.
Step 2: Derivation.
$\omega L = \frac{1}{\omega C} \Rightarrow \omega^2 = \frac{1}{LC} \Rightarrow \omega = \frac{1}{\sqrt{LC}}$
$f_r = \frac{1}{2\pi \sqrt{LC}}$.
Step 3: Application.
Tuning mechanisms in radios and TVs to catch a specific broadcast frequency while rejecting others.
Step 1: Definition.
Resonance occurs when the current amplitude in an LCR circuit becomes maximum, which happens when $X_L = X_C$.
Step 2: Derivation.
$\omega L = \frac{1}{\omega C} \Rightarrow \omega^2 = \frac{1}{LC} \Rightarrow \omega = \frac{1}{\sqrt{LC}}$
$f_r = \frac{1}{2\pi \sqrt{LC}}$.
Step 3: Application.
Tuning mechanisms in radios and TVs to catch a specific broadcast frequency while rejecting others.
Case-Based Questions
11-13. Case Study: Power Transmission
High voltage transmission lines are used to transport electrical energy from power plants to cities. A step-up transformer increases the voltage before transmission, and step-down transformers reduce it near our homes.
Q11. Why is electricity transmitted at high voltages?
Answer: To minimize power loss in transmission cables. Power $P = VI$. For a given power, higher voltage means lower current. Since line loss is $I^2R$, lowering current drastically reduces heat loss.
Q12. What type of transformer is used at the generating station?
Answer: Step-up transformer (increases V, decreases I).
Q13. Can a transformer step up AC power?
Answer: No. A transformer can step up voltage, but it cannot step up power. Power output is ideally equal to power input ($P_{in} = P_{out}$).
Answer: To minimize power loss in transmission cables. Power $P = VI$. For a given power, higher voltage means lower current. Since line loss is $I^2R$, lowering current drastically reduces heat loss.
Q12. What type of transformer is used at the generating station?
Answer: Step-up transformer (increases V, decreases I).
Q13. Can a transformer step up AC power?
Answer: No. A transformer can step up voltage, but it cannot step up power. Power output is ideally equal to power input ($P_{in} = P_{out}$).
Assertion-Reason Questions
(A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
14. Assertion (A): In a purely capacitive circuit, the current leads the voltage by 90°. Reason (R): The capacitive reactance is directly proportional to the frequency of AC.
Answer: (C).
Explanation: Assertion is true. Reason is false because $X_C = \frac{1}{2\pi fC}$, so it is inversely proportional to frequency.
Explanation: Assertion is true. Reason is false because $X_C = \frac{1}{2\pi fC}$, so it is inversely proportional to frequency.
15. Assertion (A): A 220 V AC is more dangerous than a 220 V DC. Reason (R): The peak voltage of a 220 V AC is roughly 311 V, which is much higher than the constant 220 V of DC.
Answer: (A).
Explanation: The reason correctly explains the assertion.
Explanation: The reason correctly explains the assertion.
Common Mistakes Students Make
- Confusing Peak and RMS Values: Examiners usually give the RMS value in questions (e.g., "220 V supply"). Students often treat this as Peak Voltage ($V_m$) in formulas. Remember, household voltage is always RMS!
- Wrong Phase Angles: Always remember "CIVIL": In a Capacitor, I (current) leads Voltage. Voltage leads I (current) in an L (inductor).
- Forgetting to Convert Units: Inductance given in mH (millihenry) or µF (microfarad) must be converted to standard H and F using $10^{-3}$ and $10^{-6}$ respectively before calculating reactance.
- Square Root Calculation in Impedance: Students mess up the math in $Z = \sqrt{R^2 + (X_L - X_C)^2}$. Solve the bracket first, square it, add to $R^2$, and then take the square root.
Exam Preparation Tips
- Derivations are Key: Practice the phasor diagrams and derivations for average power ($P_{avg} = V_{rms} I_{rms} \cos\phi$) and LCR impedance at least 3 times.
- Formula Sheet: Make a cheat sheet for all the formulas ($X_L, X_C, Z, f_r, Q, P$). Roz revise karo!
- Solve PYQs: Alternate Current is highly predictable. Practice previous year board questions, especially numericals on resonance and transformers.
- Time Management: In the exam, draw neat phasor diagrams with a pencil. It instantly gives a good impression and can earn you partial marks even if your final calculation goes slightly wrong.
FAQ Section
Is Chapter 7 Alternating Current important for CBSE Class 12 boards?
Yes, it is very important. Together with Electromagnetic Induction (EMI), it carries significant weightage (around 8 marks), often featuring a 3-mark or 5-mark long answer question (usually LCR circuit or Transformer).
What is the most important topic in Alternating Current Class 12?
The Series LCR circuit, condition of resonance, phasor diagrams, and Transformers are the most frequently asked topics in board exams.
Can I score full marks in numericals of this chapter?
Absolutely. The numericals are direct formula-based. Just ensure your unit conversions (like µF to F) are correct and your calculation is neat.
What is the difference between an ideal and a practical transformer?
An ideal transformer has 100% efficiency (no power loss). A practical transformer has losses due to flux leakage, copper resistance, and eddy currents, making its efficiency less than 100%.
Where can I download NCERT Solutions for Class 12 Physics Chapter 7 in PDF?
You can save this page or print it directly using your browser to use it as a complete PDF guide for Chapter 7 notes and NCERT solutions.
Mastering Chapter 7 Alternating Current is all about visualizing phasor diagrams and understanding how L, C, and R behave differently under an AC source. Don't skip the derivations and ensure you practice the numericals by hand. Save this post, practice the Board Style Important Questions, and you will be well-prepared for your 2026 Physics exams. Keep studying hard, tension mat lo, and good luck with your revisions!
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