# Alternating Current Class 12 notes Physics Chapter 7

## Introduction

Currents whose direction does not change with time through a load, are known as **direct current** (D.C.), whereas currents whose direction changes periodically through a load are known as **alternating currents** (A.C.) and the voltage is known as **alternating voltage** (ac voltage).

Electrical energy in a.c. form can be easily transmitted over long distances without much loss. A.C. voltage can be easily converted to other voltages by step up/step down **transformers**.

## Alternating Current

An **alternating current** changes its direction of flow periodically. For a half cycle, it flows in one direction and for the next half cycle, it flows in opposite direction.

### Mean value for half-cycle of AC

Mean value of AC is the total charge that flows through a circuit element in a given time interval divided by the time interval.

`I_{mean}=\frac{\int_{0}^{T}Idt}{T}`

For half cycle

`I_{mean}=\frac{\int_{0}^{T/2}Idt}{T/2}`

`I_{mean}=\frac{2}{T}\int_{0}^{T/2}I_{0}sin\omega dt`

`I_{mean}=\frac{2I_{0}}{T}[\frac{-cos\omega t}{\omega}]_{0}^{T/2}`

`I_{mean}=\frac{2I_{0}}{2\pi}[-cos\pi-cos0]` .....(`\because\omega=\frac{2\pi}{T}`)

`I_{mean}=\frac{2I_{0}}{\pi}`

**Note:** For complete cycle, mean value = 0

### RMS value of alternating supply

`I_{rms}^{2}=\frac{1}{T}\int_{0}^{T}I^{2}dt`

`I_{rms}^{2}=\frac{1}{T}\int_{0}^{T}I_{0}^{2}sin^{2}\omega dt`

`I_{rms}^{2}=\frac{1}{T}\int_{0}^{T}\frac{I_{0}^{2}(1-cos2\omega tdt)}{2}`

`I_{rms}^{2}=\frac{I_{0}^{2}}{2T}[T-\frac{[sin2\omega t]_{0}^{T}}{2\omega}]`

`I_{rms}^{2}=\frac{I_{0}^{2}}{2T}[T-\frac{(sin2\omega T-sin0)}{2\omega}]`

`I_{rms}=\frac{I_{0}}{\sqrt{2}}`

## Recommended Books

- NCERT Textbook For Class 12 Physics Part 1 & 2
- CBSE All In One Physics Class 12 2022-23 Edition
- Oswaal CBSE Chapterwise Question Bank Class 12 Physics Book
- Modern's abc Plus of Physics for Class-12 (Part I & II)

**Read also:** Electromagnetic Waves Class 12 Physics Notes Chapter 8

## A.C. Voltage Applied to a Resistor

V = V_{m} sinÏ‰t

The instantaneous power dissipated in the resistor is

`p=I^{2}R=I_{m}^{2}R sin^{2}\omega t`

The average value of ‘p’ over a cycle is

`\bar{p}=`<`I^{2}R`> = <`I_{m}^{2}Rsin^2\omega t`> ....<`sin^2\omega t`>`=\frac{1}{2}`

`\bar{p}=\frac{I^{2}R}{2}`

Similarly `V_{rms}=\frac{V_m}{\sqrt{2}}=0.707V_{m}`

To find the value of current through the resistor, we apply ohm's law

V = IR

`V_{m}sin\omega t=IR`

`I=\frac{V_{m}}{R}sin\omega t`

Since R is a constant, we can write this equation as

`I=I_{m}sin\omega t`

In a pure resistor, the voltage and current are in phase. The minima, zero, and maxima occur at the same respective times.

**Read also:** The p-Block Elements Class 12 Chemistry Notes Chapter 7

## Alternating Voltage Applied to an Inductor

Let the voltage across the source be V = V_{m} sinÏ‰t.

If, there be no resistor in the circuit,

`V-L\frac{dI}{dt}=0`

`V_{m} sin\omega t-L\frac{dI}{dt}=0`

`\frac{dI}{dt}=\frac{V_m}{L} sin\omega t`

`\int\frac{dI}{dt}dt=\frac{V_m}{L}\int sin\omega tdt`

`I=\frac{V_m}{L}(\frac{-cos\omega t}{\omega})`

`I=\frac{V_m}{\omega L}(-cos\omega t)` .....[ `X_L=\omega L` ]

`I=-\frac{V_m}{X_L}sin(\frac{\pi}{2}-\omega t)`

`I=I_{0}sin(\frac{\omega t-\pi}{2})`

In a pure Inductor, the source voltage and the current show that the current lags the voltage by Ï€/2.

### Inductive Reactance (X_{L})

The opposing nature of the inductor to the flow of current is called **Inductive reactance**.

`X_{L}=\omega L=2\pi fL`

Where, L = self-inductance

**Read also:** Conceptual Questions for Class 12 Physics Chapter 7 Alternating Current

## Alternating Voltage Applied to a Capacitor

Let the voltage across the source be V = V_{m} sinÏ‰t to a capacitor only, a purely capacitive ac circuit.

Let q be the charge on the capacitor at any time t. The instantaneous voltage V across the capacitor is

`V=\frac{q}{C}`

`V_m sin\omega t=\frac{q}{C}`

`q=CV_m sin\omega t`

To find the current, we divide the eq. by dt

`\frac{dq}{dt}=\frac{d}{dt}(CV_m sin\omega t)`

`I=CV_m cos\omega t.\omega`

`I=\frac{V_m}{1/{\omega C}} cos\omega t`

`I=\frac{V_m}{X_C} cos\omega t` .....[`X_C=\frac{1}{\omega C}`]

`I=I_m sin(\omega t+\pi/2)`

In a pure Inductor, the source voltage and the current show that the current is Ï€/2 ahead of the voltage.

### Capacitive Reactance (X_{C})

The opposing nature of capacitor to the flow of alternating current is called **capacitive reactance**.

`X_{C}=\frac{1}{\omega C}=\frac{1}{2\pi fC}`

Where C = capacitance

## Power in AC Circuits : The Power Factor

`V = V_{m} sin\omega t`

`I=I_{m} sin(\omega t+\phi)`

P = V × I

`P=V_{m}.I_{m}.sin\omega t.sin(\omega t+\phi)`

`P=\frac{V_{m}.I_{m}}{2}.2sin\omega t.sin(\omega t+\phi)`

`P=\frac{V_{m}.I_{m}}{2}.[cos(\omega t+\phi-\omega t)-cos(\omega t+\omega t+\phi)`

`P=\frac{V_{m}.I_{m}}{2}.[cos\phi-cos(2\omega t+\phi)`

`P=\frac{V_{m}}{\sqrt{2}}.\frac{I_{m}}{\sqrt{2}}.cos\phi`

`P=V_{rms}.I_{rms}.cos\phi` .....[`cos\phi=\frac{R}{Z}`]

As cos(2Ï‰t + Î¦) for one complete cycle, is ZERO

The term cosÎ¦ is known as the **power factor** as it determines the power consumed in the circuit.

**Case-1: Purely resistive circuit:**Î¦ = 0Âº, so cosÎ¦ = 1, the power dissipated is maximum. Thus**maximum power**consumed`PV=I_{rms} V_{rms}`

**Case-2: Purely Inductive or Capacitive Circuits:**The phase difference `\phi=\frac{\pi}{2}`. Thus power factor cosÎ¦ = 0. Thus power consumed is**zero**. Although current flows through the circuit but power consumed is zero. Such a current is known as**wattless current**.

## AC Voltage Applied to a Series LCR Circuit

If q is the charge in the capacitor and I is the current at time t, then applying Kirchhoff’s loop rule.

`L\frac{dI}{dt}+IR+\frac{q}{C}=V`

The voltage between inductor and capacitor is equal to V_{C} – V_{L}. The total voltage given by

`V=\sqrt{V_{R}^{2}+(V_{C}-V_{L})^2}`

`V=\sqrt{IR^{2}+(IX_{C}-IX_{L})^2}`

`V=I\sqrt{R^{2}+(X_{C}-X_{L})^2}`

`\frac{V}{I}=\sqrt{R^{2}+(X_{C}-X_{L})^2}`

`Z=\sqrt{R^{2}+(X_{C}-X_{L})^2}`

Z = (impedance = Effective resistance of series LCR circuit)

### Phase relationship between V and I

`tan\phi=\frac{V_{C}-V_{L}}{V_R}=\frac{X_{C}-X_{L}}{R}`

### Resonance

When X_{C} = X_{L}, it means

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

`\frac{1}{2\pi fC}=2\pi fL`

`f=\frac{1}{2\pi\sqrt{LC}}`

It is known as resonant frequency.

## Choke Coil

A **choke coil** is an inductor having a small resistance. It is a device used in ac circuits to control current without wasting too much power. As it has low resistance, its power factor `cos\phi` is low.

## LC Oscillations

Let a capacitor of capacitance C is initially charged to a charge q_{m}. This capacitor is connected to an inductor (having inductance L) at t=0 sec. Let at t=t sec, the charge upon capacitor and current through the inductor, are q(t) and I(t) are respectively,

`V_{C}=V_{L}\Rightarrow\frac{q}{C}=\frac{LdI}{dt}` .....(i)

Let the charge of the capacitor decrease by dq in the time interval from t = t sec to t = (t + dt) sec. Thus current

`I=\frac{dq}{dt}` .....(ii)

From (i) & (ii),

`\frac{q}{C}=L\frac{d}{dt}[\frac{-dq}{dt}]=-L\frac{d^{2}q}{dt^2}`

`\frac{d^{2}q}{dt^2}=\frac{-1}{LC}\times q`

`\frac{d^{2}q}{dt^2}+\frac{1}{LC}\times q=0`

It is a differential equation of 2nd order which is the equation of SHM in differential form. The LC oscillation is similar to the mechanical oscillation. For a **simple harmonic oscillator,** charge oscillates with natural frequency (Ï‰_{0})

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

## Transformers

**Transformers** are based upon **mutual induction** which transforms an alternating voltage from one to another of greater or smaller value.

**Construction:** A **transformer** consists of two coils wound on a soft iron core, called **primary** and **secondary** coils. Let the number of turns in these coils be N_{p} and N_{s} respectively. The input A.C. voltage is applied across the primary coil whereas the output A.C. voltage is across the secondary coil.

We consider an ideal transformer in which the primary has negligible resistance and all the flux in the core links both the primary and secondary windings. Let Î¦ be the flux linkage through each of the primary and secondary coils. Then.

Induced emf across the primary coil,

`\epsilon_{p}=-N_{p}\frac{d\phi}{dt}` ...(i)

Similarly induced emf across secondary,

`\epsilon_{s}=-N_{s}\frac{d\phi}{dt}` ...(ii)

From these equations,

AC voltage obtained across secondary / AC voltage applied across primary

`=\frac{V_s}{V_p}=\frac{\epsilon_s}{\epsilon_p}=\frac{N_s}{N_p}=r`

Where r is called transformation ration. In a transformer, some energy is always lost. The efficiency of a well designed transformer may be upto 95%. If the transformer is assumed to be 100% efficient

`p=I_{p}V_{p}=I_{s}V_{s}`

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

In actual transformers, small energy losses occur due to the following reasons.

**Flux leakage:**There is always some flux leakage. Not all the flux due to primary passes through the secondary.**Resistance of the windings:**Some energy is lost in the form of heat dissipation. It can be minimized using thick wire in case of high current, low voltage windings.**Eddy currents:**The alternating magnetic flux induces eddy currents in the iron core and causes heating. The loss can be minimized using a laminated iron core.**Hysteresis:**The magnetization of the core is repeatedly reversed by an alternating magnetic field. The resulting expenditure of energy in the core appears as heat and is kept to a minimum by using a material that has a low magnetic hysteresis loss.

### Use of Transformers in Transmission

**In electric power**transmission, transformers allow transmission of electric power at high voltages, which reduces the loss due to the heating of the wires.In many

**electronic devices**, a transformer is used to convert voltage from the distribution wiring to convenient values for the circuit requirements.**Signal**and**audio transformers**are used to couple stages of amplifiers and to match devices such as microphones and record players to the input of amplifiers.**Audio transformers**allowed telephone circuits to carry on a two-way conversation over a single pair of wires.**Resonant transformers**are used for coupling between stages of radio receivers, or in high-voltage Tesla coils.

## Summary

In a

**purely resistive**AC circuit, voltage and current are in the same phase.In a

**purely resistive**circuit, average power loss = I^{2}_{rms}x R.In a

**purely inductive**circuit, voltage is ahead of current by Ï€/2. In this circuit, average power loss = ZERO.In a

**purely capacitive**AC circuit, The current leads the applied voltage by Ï€/2. The average power loss per cycle is ZERO.For a given LCR circuit, The

**average power**consumed = V_{rms}× I_{rms}× cosÎ¸ where cosÎ¸ is the**power factor**.In the purely

**inductive**or**capacitive**circuit, cosÐ¤ = 0. Average power loss = 0. Although the current is flowing in the circuit. Such a current is known as**wattless current**.Phase relationship in a.c. circuits are represented by a

**phasor diagram**. A phasor is a vector that relates to the**angular velocity**Ï‰. The magnitude of the phasor is the**peak value**of voltage or current.The

**quality factor**is an indicator of the sharpness of the resonance.In a series LCR circuit, at resonance, X

_{L}= X_{C}., the impedance Z is minimum and equal to R.A

**step-up transformer**converts low ac voltage to high ac voltage but reduces the current.A

**step-down transformer**converts high ac voltage to a low ac voltage but increases the currents accordingly.240 V ac means it is the RMS value of ac voltage. The amplitude of this voltage V

_{M}= 240√2 = 340 volt.Power consumed in a circuit is never negative.

The constant value of dc which produces the same heat through a resistive element, due to the alternating current, is known as the

**root mean square value**of ac.The only element which dissipates energy in ac circuit is a

**resistor**.The

**power factor**in an LCR circuit is a measure of how close the circuit is to expanding the maximum power.A

**generator**converts mechanical energy into electrical energy whereas an electric motor converts electrical energy into mechanical energy.A

**transformer**does not violate the conservation of energy. A step-up transformer changes low voltage to high voltage but reduces the current in the same proportion.