NCERT Class 12 Physics Chapter 7 Alternating Current Notes - Download Free PDF

Alternating Current (AC) is an important and high-scoring chapter in Class 12 Physics. AC usually contributes around 6–8 marks, making it one of the most scoring topics in the syllabus.

• AC Current CBSE Weightage: 8–12 marks
• AC Current JEE Main Weightage: 6–8 marks (2–3 questions)
• AC Current JEE Advanced Weightage: 8–12 marks (2–3 questions)

You can find complete revision notes on alternating current class 12 NCERT chapter 7 including all formulas and numerical tips as noted for CBSE boards in the article below.

Collegedunia’s Alternating Current notes are based on NCERT, Reference books like S.L Arora, and previous years’ questions, which will help you not only in Boards but in Competitive exams as well.

AC Current Boards Revision Notes Vs JEE Main Revision Notes

Around 80% of the content is common between CBSE and JEE. Start by mastering the CBSE syllabus, then layer in the additional JEE material.

Important derivations in Alternating Current class 12 NCERT notes with step by step explanation

RMS (Root Mean Square) Value of Alternating Current and Voltage

The instantaneous current is given by

$$ i = i_m \sin \omega t $$

The mean square current over one period \( T \) is

$$ \langle i^2 \rangle = \frac{1}{T} \int_0^T i^2 \, dt = \frac{1}{T} \int_0^T i_m^2 \sin^2 \omega t \, dt $$

Using the identity \( \langle \sin^2 \omega t \rangle = \frac{1}{2} \),

$$ \langle i^2 \rangle = \frac{i_m^2}{2} $$

Therefore,

$$ I_{\text{rms}} = \sqrt{\langle i^2 \rangle} = \frac{i_m}{\sqrt{2}} = 0.707 \, i_m $$

Similarly,

$$ V_{\text{rms}} = \frac{v_m}{\sqrt{2}} $$

Key Result: \( I_{\text{rms}} = \frac{i_m}{\sqrt{2}} \), \( V_{\text{rms}} = \frac{v_m}{\sqrt{2}} \) (Equation 7.6 in NCERT).

2. AC Voltage Applied to a Pure Resistor

NCERT Reference: Section 7.2

Step-by-step Derivation:

Applied voltage:

$$ v = v_m \sin \omega t $$

From Ohm's law:

$$ i = \frac{v}{R} = \frac{v_m}{R} \sin \omega t = i_m \sin \omega t \quad \left( i_m = \frac{v_m}{R} \right) $$

→ Voltage and current are in phase (phase difference = 0).

Instantaneous power:

$$ p = i^2 R = i_m^2 R \sin^2 \omega t $$

Average power over one cycle:

$$ P = \langle p \rangle = i_m^2 R \langle \sin^2 \omega t \rangle = \frac{i_m^2 R}{2} = I_{\text{rms}}^2 R = V_{\text{rms}} I_{\text{rms}} $$

Key Result: Current is in phase with voltage; power is dissipated only in the resistor.

3. AC Voltage Applied to a Pure Inductor – Inductive Reactance

NCERT Reference: Section 7.4

Step-by-step Derivation:

Applied voltage:

$$ v = v_m \sin \omega t $$

Kirchhoff's loop rule:

$$ v = L \frac{di}{dt} $$

$$ v_m \sin \omega t = L \frac{di}{dt} $$

$$ \frac{di}{dt} = \frac{v_m}{L} \sin \omega t $$

Integrating:

$$ i = \int \frac{v_m}{L} \sin \omega t \, dt = -\frac{v_m}{\omega L} \cos \omega t + C $$

For steady-state AC (constant \( C = 0 \)):

$$ i = \frac{v_m}{\omega L} \sin \left( \omega t - \frac{\pi}{2} \right) = i_m \sin \left( \omega t - \frac{\pi}{2} \right) $$

where \( i_m = \frac{v_m}{\omega L} \).

Inductive reactance:

$$ X_L = \omega L = 2\pi \nu L $$

Key Result: Current lags voltage by \( \frac{\pi}{2} \); \( X_L \) opposes change in current.

4. AC Voltage Applied to a Pure Capacitor – Capacitive Reactance

NCERT Reference: Section 7.5

Step-by-step Derivation:

Applied voltage:

$$ v = v_m \sin \omega t $$

Charge on capacitor:

$$ q = Cv = Cv_m \sin \omega t $$

Instantaneous current:

$$ i = \frac{dq}{dt} = C \frac{dv}{dt} = C \omega v_m \cos \omega t = i_m \sin \left( \omega t + \frac{\pi}{2} \right) $$

where \( i_m = \omega C v_m = \frac{v_m}{X_C} \) and \( X_C = \frac{1}{\omega C} \).

Key Result: Current leads voltage by \( \frac{\pi}{2} \); \( X_C = \frac{1}{\omega C} \) (Equation 7.18 in NCERT).

5. AC Voltage Applied to Series LCR Circuit – Impedance and Phase Difference

NCERT Reference: Section 7.6 (Phasor-diagram solution)

Step-by-step Derivation (using phasors):

• \( V_R = i_m R \) (in phase with current \( I \))
• \( V_L = i_m X_L \) (leads \( I \) by \( \pi/2 \))
• \( V_C = i_m X_C \) (lags \( I \) by \( \pi/2 \))

Net voltage amplitude (phasor sum):

$$ v_m = \sqrt{V_R^2 + (V_L - V_C)^2} = i_m \sqrt{R^2 + (X_L - X_C)^2} $$

Impedance:

$$ Z = \frac{v_m}{i_m} = \sqrt{R^2 + (X_L - X_C)^2} $$

Phase angle \( \phi \) (voltage leads current by \( \phi \)):

$$ \tan \phi = \frac{X_L - X_C}{R} \quad \Rightarrow \quad \phi = \tan^{-1} \left( \frac{X_L - X_C}{R} \right) $$

Key Result: \( Z = \sqrt{R^2 + (X_L - X_C)^2} \), \( i = i_m \sin(\omega t + \phi) \) (Equation 7.26 & 7.27).

6. Resonance in Series LCR Circuit

NCERT Reference: Section 7.6.2

Step-by-step Derivation:

At resonance, \( X_L = X_C \):

$$ \omega L = \frac{1}{\omega C} $$

$$ \omega_0 = \frac{1}{\sqrt{LC}} $$

At \( \omega = \omega_0 \):

\( Z = R \) (minimum), \( i_m = \frac{v_m}{R} \) (maximum current), \( \phi = 0 \) (voltage and current in phase).

Key Result: Resonant frequency \( \nu_0 = \frac{1}{2\pi \sqrt{LC}} \).

7. Average Power in AC Circuit – Power Factor

NCERT Reference: Section 7.7

Step-by-step Derivation:

Instantaneous power:

$$ p = vi = v_m i_m \sin \omega t \, \sin(\omega t + \phi) $$

Using trig identity and averaging over one cycle:

$$ P = \frac{1}{2} v_m i_m \cos \phi = V_{\text{rms}} I_{\text{rms}} \cos \phi = I_{\text{rms}}^2 R $$

Power factor: \( \cos \phi = \frac{R}{Z} \)

Key Result: Average power \( P = VI\cos \phi \) (Equation 7.30). Power is dissipated only in the resistor; \( \cos \phi = 1 \) at resonance.

8. Transformer (Ideal)

NCERT Reference: Section 7.8

Step-by-step Derivation (from mutual induction):

For ideal transformer (no losses):

$$ \frac{V_s}{V_p} = \frac{N_s}{N_p} \quad \text{(voltage ratio)} $$

$$ \frac{I_s}{I_p} = \frac{N_p}{N_s} \quad \text{(current ratio)} $$

Power conservation: \( V_p I_p = V_s I_s \).

These derivations are directly from the NCERT textbook and form the core of the chapter. They are frequently asked as 5-mark or 3-mark questions in board exams.

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