Alternating Current

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Home Concept Boosters CBSE Class 12 Physics Alternating Current

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How to Use This Page

Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page covers Alternating Current (AC) completely for CBSE Class 12 Physics.

Key Concepts

Class 12 · Physics · Alternating Current

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Core concepts you must know

Class 12 · Ch 7

Alternating Current (AC) Formula

An electric current that periodically reverses its direction and changes its magnitude continuously with time. $\omega$ is the angular frequency and $\phi$ is the initial phase.

$$i = i_0 \sin (\omega t + \phi) \quad | \quad T = \frac{2 \pi}{\omega}$$

Mean (Average) & RMS Current Formula

The average value of AC over a complete cycle is zero. The Root Mean Square (RMS) value is the effective value of AC that produces the same heating effect as a steady DC current.

$$\bar{i} = \frac{1}{T} \int_0^T i \, dt = 0 \quad | \quad i_{\text{rms}} = \left[\frac{1}{T} \int_0^T i^2 \, dt\right]^{1/2} = \frac{i_0}{\sqrt{2}}$$

Reactance (Inductive & Capacitive) Formula

The opposition offered to the flow of AC. Inductive reactance ($X_L$) increases with frequency, while capacitive reactance ($X_C$) decreases with frequency. SI unit is Ohm ($\Omega$).

$$X_L = \omega L \quad | \quad X_C = \frac{1}{\omega C}$$

Impedance ($Z$) & Phase Angle ($\phi$) Formula

Impedance is the total effective opposition offered by a combination of $R$, $L$, and $C$ to AC. The phase angle tells us how much the current leads or lags the voltage.

$$Z = \frac{e_0}{i_0} = \frac{e_{\text{rms}}}{i_{\text{rms}}}$$

$$\text{LR Circuit: } Z = \sqrt{R^2 + \omega^2 L^2}, \quad \tan \phi = \frac{\omega L}{R}$$

$$\text{RC Circuit: } Z = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}, \quad \tan \phi = \frac{1}{\omega C R}$$

LCR Series Circuit & Resonance Formula

When $X_L = X_C$, their opposing effects cancel out. Impedance becomes minimum ($Z = R$), and current becomes maximum. This specific frequency is the resonance frequency.

$$Z = \sqrt{R^2 + \left(\omega L – \frac{1}{\omega C}\right)^2}, \quad \tan \phi = \frac{\omega L – \frac{1}{\omega C}}{R}$$

$$\nu_{\text{resonance}} = \frac{1}{2 \pi \sqrt{LC}}$$

AC Power & Power Factor Formula

The true power consumed in an AC circuit depends not just on voltage and current, but also on the cosine of the phase angle, known as the Power Factor ($\cos \phi$).

$$P = e_{\text{rms}} i_{\text{rms}} \cos \phi \quad | \quad E_{\text{energy}} = i_{\text{rms}}^2 R T$$

Transformer Principle Formula

Works on mutual induction. It steps up or steps down alternating voltage. For an ideal transformer, input power equals output power.

$$\frac{N_1}{N_2} = \frac{e_1}{e_2} = \frac{i_2}{i_1} \quad | \quad e_1 i_1 = e_2 i_2$$

Concept Deep Dive

The Magic of Resonance

How tuning your radio actually works

Core Concept

When you turn the knob on a radio, you are changing the capacitance ($C$) of a variable capacitor inside an LCR circuit. The antenna receives thousands of radio frequencies simultaneously from different stations. By changing $C$, you change the natural resonance frequency of your radio’s circuit ($\nu_r = 1 / 2\pi\sqrt{LC}$).

When this matches the frequency of a specific station (say 98.3 MHz), $X_L$ perfectly cancels $X_C$. The impedance drops to its lowest possible value ($Z=R$), and a massive current from that specific station surges through the circuit, overpowering all other stations.

$$\text{At Resonance: } X_L = X_C \implies Z = R \implies I_{\text{max}} = \frac{V}{R}$$

Wattless Current

Current that does no work

High Yield

In a purely inductive ($L$) or purely capacitive ($C$) circuit, the phase difference between voltage and current is exactly $90^\circ$ ($\pi/2$). The power consumed is $P = e_{\text{rms}} i_{\text{rms}} \cos(90^\circ) = 0$.

This means a massive current could be surging back and forth in the circuit, but the net power consumed over a full cycle is absolutely zero! The energy simply sloshes back and forth between the source and the magnetic/electric field of the component. The component of current causing this ($i_{\text{rms}} \sin \phi$) is called Wattless Current.

Compare & Contrast

✗ Inductor in AC ($L$)

Reactance $X_L = \omega L = 2\pi\nu L$.
Reactance is directly proportional to frequency ($\nu$).
Blocks High-Frequency AC (High $\nu$ = High Resistance).
Allows DC to pass easily ($\nu = 0 \implies X_L = 0$).

✓ Capacitor in AC ($C$)

Reactance $X_C = \frac{1}{\omega C} = \frac{1}{2\pi\nu C}$.
Reactance is inversely proportional to frequency ($\nu$).
Blocks DC completely ($\nu = 0 \implies X_C = \infty$).
Allows High-Frequency AC to pass easily.

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Remember

Because of these exact opposing properties, we use Inductors as “Choke Coils” to reduce high-frequency AC current without wasting power as heat (unlike resistors). We use Capacitors to filter out DC components from mixed signals.

Common Mistakes to Avoid

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Mistake 1

Algebraic Addition of Voltages: In an LCR circuit, if you measure $30\text{V}$ across the resistor, $40\text{V}$ across the inductor, and $20\text{V}$ across the capacitor, the total supply voltage is NOT $30+40+20 = 90\text{V}$. You MUST use phasor addition: $V = \sqrt{V_R^2 + (V_L – V_C)^2} = \sqrt{30^2 + (40-20)^2} = \sqrt{900+400} \approx 36\text{V}$.

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Mistake 2

Confusing $\omega$ and $\nu$: In the reactance formulas, $\omega$ is the angular frequency ($2\pi\nu$) measured in rad/s. $\nu$ (or $f$) is the linear frequency measured in Hertz (Hz). If a question says “50 Hz AC supply”, you must multiply by $2\pi$ to get $\omega$ before calculating $X_L$ or $X_C$.

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Mistake 3

Misinterpreting Meter Readings: Ordinary DC ammeters and voltmeters read zero in an AC circuit because the average value of AC is zero. Specialized AC meters (hot-wire instruments) measure the heating effect, which means they always read the RMS value. If an AC voltmeter reads $220\text{V}$, that is $V_{\text{rms}}$, not the peak voltage $V_0$!

Exam Tips

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Tip 1

To quickly find the Power Factor ($\cos \phi$) from an impedance triangle, use the formula $\cos \phi = \frac{R}{Z}$. At resonance, since $Z = R$, the power factor is exactly $1$ (maximum possible power consumption).

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Tip 2

In transformer numericals, an “ideal” transformer is assumed to be $100\%$ efficient ($\eta = 1$), meaning Input Power = Output Power ($V_p I_p = V_s I_s$). If an efficiency is given (e.g., $90\%$), you must use $\eta = \frac{V_s I_s}{V_p I_p}$ to find the currents.

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Did You Know

Nikola Tesla championed AC because it could be easily stepped up to very high voltages (like $400,000\text{V}$) using transformers for long-distance transmission. High voltage means low current for the same power, which drastically reduces $I^2R$ heat losses in the power lines!

Expected Exam Questions

Board Pattern Questions

Class 12 · Alternating Current · CBSE Exam

Class 12 · Physics

Why is a $220\text{V}$ AC supply more dangerous than a $220\text{V}$ DC supply? [1 mark]

Answer Because AC reaches a peak voltage of $311\text{V}$. 📝

Explanation

The stated value of an AC supply ($220\text{V}$) is its RMS value ($V_{\text{rms}}$). The peak voltage it reaches during every cycle is $V_0 = \sqrt{2} \times V_{\text{rms}} = 1.414 \times 220 \approx 311\text{V}$. A $220\text{V}$ DC supply stays constantly at $220\text{V}$. Thus, the AC supply exposes a person to a much higher maximum voltage, making it more lethal.

A step-down transformer is used to reduce the main supply of $220\text{V}$ to $11\text{V}$. If the primary draws a current of $5\text{A}$ and the secondary draws $90\text{A}$, what is the efficiency of the transformer? [2 marks]

Answer $90\%$ 📝

Explanation

Given: $V_p = 220\text{V}$, $I_p = 5\text{A}$, $V_s = 11\text{V}$, $I_s = 90\text{A}$.
Input Power $P_{in} = V_p I_p = 220 \times 5 = 1100\text{W}$.
Output Power $P_{out} = V_s I_s = 11 \times 90 = 990\text{W}$.
Efficiency $\eta = \left(\frac{P_{out}}{P_{in}}\right) \times 100\% = \left(\frac{990}{1100}\right) \times 100\% = 0.9 \times 100\% = 90\%$.

In a series LCR circuit, the voltage across the inductor ($V_L$) is $40\text{V}$, across the capacitor ($V_C$) is $10\text{V}$, and across the resistor ($V_R$) is $40\text{V}$. Calculate the impedance of the circuit if the current flowing is $2\text{A}$. [3 marks]

Answer $25\,\Omega$ 📝

Explanation

First, find the total effective RMS voltage using phasor addition:
$V = \sqrt{V_R^2 + (V_L – V_C)^2}$
$V = \sqrt{40^2 + (40 – 10)^2} = \sqrt{1600 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50\text{V}$.
According to Ohm’s law for AC, Impedance $Z = V / I$.
$Z = \frac{50\text{V}}{2\text{A}} = 25\,\Omega$.

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