Alternating Current Class 12 Notes for IIT JEE | NEET

Alternating Current is Chapter 7 of NCERT Class 12 Physics. In the CBSE 2026-27 scheme, Chapters 4 to 7 (Moving Charges and Magnetism, Magnetism and Matter, Electromagnetic Induction and Alternating Current) carry 17 marks together. In the JEE Main 2026 syllabus, the chapter is part of Unit 14, together with Electromagnetic Induction. These Alternating Current Class 12 Notes follow the NCERT chapter, and they give the formula and the reason for each result.

Alternating Current Class 12 Notes: Chapter Overview and Syllabus

Check the latest syllabi of CBSE and NTA, and the NMC syllabus for NEET, for the exact lists.

Tip from the eSaral Physics team: Always draw the phasor diagram before you write a formula for an AC circuit. If you know which of $X_L$ and $X_C$ is larger, you know the sign of the phase angle and whether the circuit is inductive or capacitive.

Alternating Current Class 12 Notes

India's Best Exam Preparation for Class 12th - Download Now

India's Best Exam Preparation for Class 12th - Download Now

India's Best Exam Preparation for Class 12th - Download Now

India's Best Exam Preparation for Class 12th - Download Now

India's Best Exam Preparation for Class 12th - Download Now

India's Best Exam Preparation for Class 12th - Download Now

India's Best Exam Preparation for Class 12th - Download Now

India's Best Exam Preparation for Class 12th - Download Now

What Is Alternating Current? in Alternating Current Class 12 Notes

An alternating current changes its magnitude continuously with time, and it reverses its direction periodically. A sinusoidal alternating voltage and current are:

$$v = V_0\sin\omega t, \qquad i = I_0\sin(\omega t + \phi)$$

where $V_0$ and $I_0$ are the peak values, $\omega = 2\pi f$ is the angular frequency, $T = \dfrac{1}{f}$ is the time period, and $\phi$ is the phase difference between the current and the voltage. The household supply in India is 230 V (RMS) at 50 Hz. AC is used for power supply because its voltage can be changed easily with a transformer, and it can be sent over long distances with less loss.

Peak, RMS and Mean Values in Alternating Current Class 12 Notes

The average of a sinusoidal current over a full cycle is zero, so we use the RMS (root mean square) value, which is the DC value that gives the same heating in a resistor:

$$I_{rms} = \frac{I_0}{\sqrt{2}} \approx 0.707\,I_0, \qquad V_{rms} = \frac{V_0}{\sqrt{2}}$$

AC ammeters and voltmeters read RMS values, and the rating of a household supply (230 V) is an RMS value.

AC Through a Resistor, an Inductor and a Capacitor in Alternating Current Class 12 Notes

For each element, $I_0 = \dfrac{V_0}{R}$, $\dfrac{V_0}{X_L}$ or $\dfrac{V_0}{X_C}$, and the unit of reactance is the ohm. An inductor passes DC easily and opposes high frequencies, because $X_L$ rises with $f$. A capacitor blocks DC and passes high frequencies, because $X_C$ falls with $f$. A memory aid is “CIVIL”: in a capacitor (C), the current (I) leads the voltage (V), and in an inductor (L), the voltage leads the current.

Series LCR Circuit and Impedance in Alternating Current Class 12 Notes

In a series LCR circuit, the same current flows through R, L and C. The voltages $V_R$, $V_L$ and $V_C$ are added as phasors, with $V_L$ and $V_C$ opposite to each other:

$$V_0^{2} = V_R^{2} + (V_L - V_C)^{2}$$ $$Z = \sqrt{R^{2} + (X_L - X_C)^{2}}, \qquad I_0 = \frac{V_0}{Z}, \qquad \tan\phi = \frac{X_L - X_C}{R}$$

Resonance and Quality Factor in Alternating Current Class 12 Notes

At resonance, $X_L = X_C$, so the impedance is minimum ($Z = R$), the current is maximum, and the phase angle is zero:

$$\omega_0 = \frac{1}{\sqrt{LC}}, \qquad f_0 = \frac{1}{2\pi\sqrt{LC}}, \qquad I_{max} = \frac{V}{R}$$

The quality factor measures how sharp the resonance is:

$$Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R} = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{\omega_0}{2\Delta\omega}$$

where $2\Delta\omega = \dfrac{R}{L}$ is the bandwidth. A smaller resistance gives a larger $Q$, a sharper peak and a narrower bandwidth. A radio tuner uses resonance to pick one station.

Power in an AC Circuit in Alternating Current Class 12 Notes

The average power over a cycle is:

$$P = V_{rms}\,I_{rms}\cos\phi, \qquad \cos\phi = \frac{R}{Z} \ \text{(power factor)}$$

• For a purely resistive circuit, $\phi = 0$ and $\cos\phi = 1$, so $P = V_{rms}I_{rms}$.
• For a pure inductor or a pure capacitor, $\phi = 90^\circ$ and $\cos\phi = 0$, so the average power is zero. The current in this case is called the wattless current, because it does not consume power.
• At resonance, $\cos\phi = 1$, so the power is maximum.
• A low power factor means that the circuit draws a larger current for the same useful power. A capacitor in parallel is used to improve it.

LC Oscillations in Alternating Current Class 12 Notes

When a charged capacitor is connected to an inductor, the energy moves between the electric field of the capacitor and the magnetic field of the inductor. The oscillation has the angular frequency $\omega_0 = \dfrac{1}{\sqrt{LC}}$, and the total energy stays constant (in the ideal case without resistance):

$$U = \frac{q^{2}}{2C} + \frac{1}{2}LI^{2} = \text{constant}$$

For CBSE, this topic was qualitative only in earlier syllabi, so check the latest CBSE syllabus.

AC Generator and Transformer in Alternating Current Class 12 Notes

AC Generator

A coil of $N$ turns and area $A$ rotating with the angular speed $\omega$ in a uniform field $B$ produces an emf by electromagnetic induction:

$$\varepsilon = NBA\omega\sin\omega t, \qquad \varepsilon_0 = NBA\omega$$

Transformer

A transformer works on mutual induction and changes the AC voltage. For an ideal transformer, with $N_p$ and $N_s$ turns in the primary and the secondary:

$$\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}$$

Efficiency: $\eta = \dfrac{P_{out}}{P_{in}}$, which is less than 1 for a real transformer. Losses and their remedies: copper loss ($I^2R$ in the windings, reduced by thick wire), eddy current loss (reduced by a laminated core), hysteresis loss (reduced by a soft iron core) and flux leakage (reduced by winding the coils over each other).

Key Formulas in Alternating Current Class 12 Notes

Solved Examples from Alternating Current Class 12 Notes

Example 1 (peak and RMS): The household supply is 230 V (RMS). Find the peak voltage.

$V_0 = \sqrt{2}\,V_{rms} = 1.414 \times 230 \approx 325$ V.

Example 2 (reactances): Find $X_L$ for $L = 0.1$ H and $X_C$ for $C = 100\ \mu$F, both at 50 Hz.

$X_L = 2\pi f L = 2\pi \times 50 \times 0.1 \approx 31.4\ \Omega$. $X_C = \dfrac{1}{2\pi f C} = \dfrac{1}{2\pi \times 50 \times 100 \times 10^{-6}} \approx 31.8\ \Omega$.

Example 3 (series LCR circuit): In a series LCR circuit, $R = 30\ \Omega$, $X_L = 80\ \Omega$ and $X_C = 40\ \Omega$, and the supply is 100 V (RMS). Find the impedance, the current, the phase angle and the average power.

$Z = \sqrt{30^2 + (80 - 40)^2} = 50\ \Omega$. $I_{rms} = \dfrac{100}{50} = 2$ A. $\tan\phi = \dfrac{40}{30}$, so $\phi \approx 53^\circ$ and $\cos\phi = 0.6$. $P = 100 \times 2 \times 0.6 = 120$ W, which equals $I_{rms}^2 R = 4 \times 30$.

Example 4 (resonance and Q): A series circuit has $L = 1$ mH, $C = 10\ \mu$F and $R = 2\ \Omega$, with a 10 V (RMS) source. Find the resonant frequency, the current at resonance and $Q$.

$f_0 = \dfrac{1}{2\pi\sqrt{10^{-3} \times 10^{-5}}} = \dfrac{1}{2\pi \times 10^{-4}} \approx 1592$ Hz. $I = \dfrac{V}{R} = 5$ A. $Q = \dfrac{1}{R}\sqrt{\dfrac{L}{C}} = \dfrac{1}{2}\sqrt{100} = 5$.

Example 5 (power factor): A circuit draws 5 A from a 220 V supply, and the current lags the voltage by $60^\circ$. Find the average power.

$P = V_{rms}I_{rms}\cos\phi = 220 \times 5 \times 0.5 = 550$ W.

Example 6 (transformer): A step-down transformer converts 2200 V to 220 V. The primary has 1000 turns, and the secondary current is 10 A. Find the number of secondary turns, and the primary current if the efficiency is 90%.

$N_s = N_p\dfrac{V_s}{V_p} = 1000 \times \dfrac{220}{2200} = 100$. $P_{out} = 220 \times 10 = 2200$ W, so $P_{in} = \dfrac{2200}{0.9} \approx 2444$ W, and $I_p = \dfrac{2444}{2200} \approx 1.11$ A.

Example 7 (RMS and mean values): The peak current of an AC is 10 A. Find the RMS value and the mean value over a half cycle.

$I_{rms} = \dfrac{10}{\sqrt{2}} \approx 7.07$ A. $\langle I\rangle = \dfrac{2 \times 10}{\pi} \approx 6.37$ A.

Example 8 (pure inductor): An inductor of 0.5 H is connected to a 220 V, 50 Hz supply. Find the RMS current, the phase relation and the average power.

$X_L = 2\pi \times 50 \times 0.5 \approx 157\ \Omega$, so $I_{rms} = \dfrac{220}{157} \approx 1.40$ A. The current lags the voltage by $90^\circ$, and the average power is zero.

Example 9 (AC generator): A coil has 100 turns and an area of 0.05 m². It rotates at 50 Hz in a field of 0.2 T. Find the peak and the RMS emf.

$\varepsilon_0 = NBA\omega = 100 \times 0.2 \times 0.05 \times 2\pi \times 50 \approx 314$ V. $\varepsilon_{rms} = \dfrac{314}{\sqrt{2}} \approx 222$ V.

Common Mistakes in Alternating Current Class 12 Notes Problems

How to Study Alternating Current Class 12 Notes

1. Read the NCERT chapter first, and download the textbook from eSaral’s NCERT Class 12 Physics book page.
2. Learn the table of R, L and C (reactance and phase), and then practise the phasor diagram of the series LCR circuit.
3. Solve one problem each on RMS values, impedance, resonance and Q, power factor and the transformer.
4. Use eSaral’s Complete Class 12 Physics notes for all the chapters, and the free Physics revision series on YouTube for quick formula revision.
5. Revise the unit with eSaral’s JEE Main Physics revision notes (Unit 14), and test yourself with eSaral’s JEE Main chapter-wise PYQs and NEET chapter-wise PYQs.