# Alternating Current Class 12 notes Physics Chapter 7

Introduction, Alternating Current, A.C. Voltage, EMF, Power in AC Circuits, AC Voltage Applied to a Series LCR Circuit, LC Oscillations, Transformers

## 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}}

## A.C. Voltage Applied to a Resistor

V = Vm sinωt

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.

## Alternating Voltage Applied to an Inductor

Let the voltage across the source be V = Vm sinωt.

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 π/2.

### Inductive Reactance (XL)

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

## Alternating Voltage Applied to a Capacitor

Let the voltage across the source be V = Vm sinωt 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 π/2 ahead of the voltage.

### Capacitive Reactance (XC)

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(2ωt + Φ) for one complete cycle, is ZERO

The term cosΦ is known as the power factor as it determines the power consumed in the circuit.

1. Case-1: Purely resistive circuit: Φ = 0º, so cosΦ = 1, the power dissipated is maximum. Thus maximum power consumed

PV=I_{rms} V_{rms}

2. Case-2: Purely Inductive or Capacitive Circuits: The phase difference \phi=\frac{\pi}{2}. Thus power factor cosΦ = 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 VC – VL. 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 XC = XL, 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 qm. 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 (ω0)

\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 Np and Ns 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 Φ 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.

1. Flux leakage: There is always some flux leakage. Not all the flux due to primary passes through the secondary.

2. 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.

3. 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.

4. 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

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

2. In many electronic devices, a transformer is used to convert voltage from the distribution wiring to convenient values for the circuit requirements.

3. 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.

4. Audio transformers allowed telephone circuits to carry on a two-way conversation over a single pair of wires.

5. 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 = I2rms x R.

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

• In a purely capacitive AC circuit, The current leads the applied voltage by π/2. The average power loss per cycle is ZERO.

• For a given LCR circuit, The average power consumed = Vrms × Irms × cosθ where cosθ is the power factor.

• In the purely inductive or capacitive circuit, cosФ = 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 ω. 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, XL = XC., 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 VM = 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.