CBSE Class 12 Physics Revision Notes Chapter 7

Introduction to Class 12 Physics

Class 12 is an important  stepping stone  in  a student’s academic career. . The level of difficulty increases in this standard. To get admission to recognised universities and colleges, it is important for students to  get high scores in Class 12 board exams. Therefore, building a strong foundation  in Class 12 Physics is very crucial for future learning and a bright career prospects.. 

The NCERT book for Class 12th physics is an excellent book. It explains all the 15 chapters clearly and thoroughly. To gain expertise in physics, students must consider referring to various reference books or notes, such as the Class 12 Physics Chapter 7 notes provided by Extramarks. 

Read the full article to gain information about Chapter Seven- Alternating Current of Class 12 physics.  

Class 12 Physics Chapter 7 notes

The notes of Class 12 Physics Chapter 7 prepare the students not only for the Class 12 board exams but also for various state level and national level competitive exams such as NEET, JEE, etc. It assists students in learning more about the exam pattern, marking system and weightage of the chapter. The Class 12 Physics Chapter 7 notes provide accurate and apt information for students. Visit the Extramarks web portal or mobile app to get access to various academic notes and enhance your exam preparation.  

Students will learn about AC voltage and alternating current in a pre resistor, pure inductor, pure capacitor and LCR circuit. By regularly practicing the problems based on this chapter, students will learn to tackle the complex questions and prepare for the board exams. Special properties of AC circuits, devices which work on alternating current, the difference between AC and Dc voltage, resonance and resonating frequency are some of the most important concepts from the Class 12 Physics Chapter 7 notes. 

 A glance through  all the key topics students will find in the CBSE Class 12 physics Chapter 7 notes. Students may refer to these revision notes to get access to various important questions, derivations, and most-likely problems for extensive practice. 

 NCERT Class 12 Physics Chapter 7 Notes: Key Topics

Topics explained in the Class 12 Physics Chapter 7 notes are:

• Introduction
• Application of AC voltage to a resistor
• Representation of AC current and voltage (Phasors)
• Application of AC voltage to an inductor
• Application of AC voltage to a capacitor
• Application of AC voltage to a series LCR circuit
• Power in AC circuit
• LC Oscillations
• Transformers

INTRODUCTION

Using the Class 12 Physics Chapter 7 Notes, students will gain information regarding the alternating current and their applications in our regular life. A number of the applications of AC current which might be seen in our day to day life are:- Radio, refrigerators, television, transformers, cars and so on. This chapter will also help to understand the implementation of AC current in these electronic appliances. Topics like resonance, LC oscillations, AC voltage and Transformers are explained in detail in the Class 12 Physics Chapter 7 Notes.  

ALTERNATING CURRENT

An alternating voltage varies like a sine function with respect to time, and the alternating current is the current driven by it. The value of the current oscillates between that maximum and minimum value.

In AC, as the value of time changes, the magnitude of current also changes. The direction is clockwise and anticlockwise, and it is repetitive. 

The frequency of AC is how fast the electrons change their directions. For e.g., if the frequency is 30 Hz, it means that the electrons move back and forth 30 times in one second. 

Direct voltage gives rise to Direct current, and similarly, Alternating voltage gives rise to Alternating current. 

AC is expressed as I = Im sinωt, where Im is the maximum value of AC. 

AC Voltage is given as V = Vm sinωt, where Vm is the maximum value of the voltage and ω is the angular frequency. 

Causes of Alternating Current:

In a rotating magnetic field, both poles of the magnet keep changing. This results in the direction of electrons being reversed and the oscillation of electrons to give rise to current. This current is called Alternating current. 

The difference between AC and DC current is given in the table below:

APPLICATION OF AC VOLTAGE TO A RESISTOR ONLY 

In a resistive AC circuit, no other elements except the resistors will be present. Consider the circuit in the figure given below.

The Input AC Voltage (V) = Vm sin ωt, where Vm is the maximum value of the voltage (voltage amplitude), and ω is the angular frequency. 

Using Kirchhoff’s loop law to the given circuit, we find the total EMF of the voltage source. 

It is given as V= IR

Therefore, Vm sin ωt = IR

I = (Vm R)sin ωt is the current which flows in the circuit. 

We can write this equation as I = Im sinωt, where Im=Vm R is the maximum value of AC and R is a constant. 

Im=Vm R is also known as Ohm’s law. 

Conclusion:

When an alternating voltage is applied to a given circuit having resistors only, then the current passing through it will be an alternating current, i.e., it will be a sinusoidal function. 

Graphical Representation:

Image Name: Representation of voltage & current

Image link: https://www.examfear.com/u-img/00/00/51/00005123.jpg

The graph plotted voltage across the resistor with Current (I) passing through it as a function of time shows that both voltage and current are sinusoidal functions. The maximum current value is less than that of the maximum voltage value. Both voltage and current are in phase with each other, which means they reach their maximum values and minimum values at the same time. 

Power of Resistor:

The average power associated with the resistor is not zero over one complete cycle. The dissipated power is in the form of heat. Using Joule’s law, heat is given as i2R. Since i is positive, we can say that the average power consumption is by pure resistive circuits. 

Using, I = Im sinωt, and V = Vm sinωt

Instantaneous power (p) = VI = Im sinωt Vm sinωt

or p= Im Vm sin2ωt …. (1) 

Average power (p) = <i2R> = <im2R sin2ωt>, where im2and R are constants. 

Therefore, (p) = im2R < sin2ωt>

We know that sin2ωt = 12(1– cos 2ωt), since < cos2ωt > = 0 

< sin2ωt> = 12

Thus, we can say that (p) = 12im2R 

To represent the expression of power in AC power in the same form as DC power, a new term for the current was introduced as the root mean square (RMS) current, denoted by Irms.

By replacing the peak value with the RMS value, we get

P = 12Irms2R 

RMS voltage:

RMS voltage, also known as the effective voltage, is obtained by taking the Root Mean Squared instantaneous voltage. It supplies the same power that an equivalent DC circuit supplies to the load, P = 12Irms2R. It also means that the power supplied to the load is equivalent to a DC circuit. 

Graphically, the RMS value of AC voltage corresponds to the same amount of power in DC Voltage. 

RMS Current:

RMS current, also known as the effective current, is the same as Root Mean Squared instantaneous current. It supplies the same power as an equivalent DC circuit to the load. Irms = Im 2.

Overview of Irms , Iavg and Iinst 

1. The instantaneous value of current

With time, the instantaneous value of the current keeps increasing. Therefore, I = Im sinωt

1. RMS value

Irms = Im 2.

1. Average value

Iavg= 0, is the average value of the current that flows through the circuit in one cycle. 

PHASORS

Phasor diagrams are used to represent the voltage-current (V-I) relationship in AC circuits. It is known to be a vector that rotates at the origin with angular velocity (). The vertical axis represents the sinusoidally variable quantity. 

Consider V = Vm sinωt, the instant worth of voltage is represented by the vertical component.

The magnitude of the phasor is the peak value at that time(t).

Phasor diagrams are used to represent the difficult relationship between the voltage and currents.

The Class 12 Physics Chapter 7 Notes have unlimited problems based on this concept. Students may practice these questions and refer to the solutions given in the notes. 

APPLICATION OF AC VOLTAGE TO A INDUCTOR ONLY 

In an inductive AC circuit, no other elements except the inductors will be present. Consider the circuit in the figure given below.

The Alternating Voltage is represented as V = Vm sinωt.

The source voltage (V) and inductor have inductance(L) in the circuit. Since there are no resistors, there is one source of EMF, i.e., the other EMF is self-induced. 

As the current keeps changing, there is a change in the magnetic flux as well. 

According to Kirchoff’s law, (t)=0. 

Therefore, V – L didt= 0, where L is inductance. 

didt=VL=Vm sinωtL

On integrating, we get

didt=VmLsinωt (dt)

I = –VmωL(cosωt) + constant

The integration constant is zero. 

–(cosωt)= sin (ωt-2)

We have I = Imsin (ωt-2), where Im = VmωL is known as the amplitude and ωL is inductive reactance. 

XL=ωL. The inductive reactance has SI unit  Ohm (Ω). The inductive reactance is proportional to inductance and frequency. 

Conclusion

The current and voltage are not in phase with each other. They are out of phase by (/2).

Graphical Representation:

Voltage and current are represented as V = Vm sinωt and I = Im sin (ωt-2), respectively. The AC Voltage and AC currents are out of phase by (/2).

The power associated with Inductor (pL)= Im sin (ωt-2) Vm sinωt = – ImVm2sin(2ωt).

The average power is zero as sin(2ωt) for a complete cycle is zero. 

Current (I) passing through the coil enters point A and increases from 0 to a peak value. The magnetic flux lines are set, i.e., the core is magnetised. With polarity shown, voltage and current are positive. Product p is positive. Therefore, energy is absorbed from the source.

The Current (I) in a coil is positive, but it is decreasing. The core is demagnetised, and the net magnetic flux is zero at the end of the half-cycle. Voltage (V) becomes negative. Therefore, a product of voltage and current is also negative, and the Energy is returned to its source. 

The Current (I) in the coil is negative. It enters at point B and exits from A. Because of the change in direction, the polarity of the magnet also changes. The product of voltage and current is positive as they both are negative. Hence, energy is absorbed. 

Current (I) decreases and reaches zero value when a core is demagnetised and magnetic flux becomes zero. The power is negative as a voltage is positive, but the current is negative. Energy absorbed during quarter cycles 2-3 is returned to its source. 

Students may consider referring to the Class 12 Physics Chapter 7 Notes to understand this concept clearly.

APPLICATION OF AC VOLTAGE TO A CAPACITOR ONLY 

The circuit has only AC Voltage, and the only element present is the Capacitor. 

In this circuit, let the AC Voltage be applied to a. Then, the source voltage is given as V = Vm sinωt. 

Also, the capacitor gets charged and discharged continuously in each half-cycle. Therefore, the voltage (V) applied is equal to the total voltage across the plate of the capacitor. Then, we know that V = qC, where q is the charge and C is a proportional constant. 

This implies, Vm sinωt = qC

By differentiating, we get 

I = C Vm ω.cosωt

Using (cosωt)= sin (ωt-2), we get

I = C Vm ω.sin (ωt-2). 

Let Im= VmR , where R = 1ωC

I = Imsin (ωt-2) is the expression for current passing through a circuit with only a capacitor. 

We denote R as XC= 1ωC is known as the capacitive reactance.

Capacitive Reactance:

The amplitude is given as Im= VmωC. The capacitive  reactance has SI unit  Ohm (Ω). The capacitive  reactance acts as a resistance for a capacitive AC circuit. 

Graphical Representation:

Voltage lags by (/2) from the current at the time t. The current is ahead by (/2) for a capacitive AC circuit.

The average current and average voltage are zero for a complete cycle.

The power associated with Capacitor is represented as the product of V = Vm sinωt and I = Im sin (ωt-2)

Therefore power (pC)= VI = Vm Im sinωt sin (ωt-2)

We get the average power of the circuit having one capacitor as zero as the average power supplied to the capacitor is zero in one complete cycle. 

The current (I) passing through the circuit is shown in the figure. The current starts from a maximum at 0 and reaches a zero value at 1. The capacitor plate A has a positive polarity, and negative charge q builds upon the other plate B reaching a maximum at one until the current is zero. The Voltage V = qC is in phase with charge q and reaches a maximum at 1. Power = VI is positive as current and voltage are positive. The capacitor is charged, so energy is absorbed from the source during this quarter.

The Current (I) now reverses the direction. The charge which was accumulated is depleted as the capacitor is discharged. The voltage reduces but is positive. The current is negative, and so power becomes negative. Energy absorbed in this cycle is returned during this quarter. The current is leading here.  

As current (I) flows from point A to B, the capacitor gets charged to reversed polarity. The plate B of the capacitor acquires positive, whereas plate A becomes negative. Since both current and voltage are negative, power p is positive. The capacitor, in this cycle, absorbs energy. 

The current (I) again reverses the direction and passes from points B to A. the charge which is accumulated is depleted, and voltage is reduced. At one point, the collage becomes zero, and the capacitor is fully discharged. Therefore, power is negative. Energy absorbed during this cycle is returned to the source.

The Class 12 Physics Chapter 7 Notes includes a detailed explanation of this chapter for full-fledged examination preparations. 

APPLICATION OF AC VOLTAGE TO LCR CIRCUIT

The LCR circuit consists of an inductor with inductance (L), a capacitor with capacitance (c) and a resistor with resistance (R). 

All the three elements in an AC circuit are connected in series, and hence, the current will flow through them. 

Let us take V = Vm sinωt

Using the Kirchhoff’s law we get,

Total EMF in a circuit = V + e = IR + qC, where

V is source voltage, 

E is self- induced emf

IR is the voltage drop across the resistor. 

q is charge

C is constant. 

Therefore, we can say that Vm sinωt – L (didt) = IR + qC

By rearranging and solving, we get

L(d2qdt2) + R (dqdt) + (qC)) = Vm sinωt

Impedance:

In an LCR circuit, the resistance with series is known as impedance. It is defined as the total (net) resistance that is offered by the LCR, i.e. the inductor, capacitor, and resistor in the circuit. Notation is Z, and the SI unit is the Ohm (Ω). 

Let the resistor, inductor and capacitor be connected in series in a circuit. The AC current is given as

I = Im sin (ωt+), where is the phase difference.

Let I be the Phasor that represents the current. Then VL, VR, VC, and V are voltages in the circuit across the elements inductor, resistor, capacitance and source, respectively. We know, VR is parallel to current (I), VC is 2 behind I and VL is 2 ahead of the Current. 

The amplitudes of VL, VR, VC are given as 

vLm= ImXL, vRm= Im and vCm= ImXC

Therefore the Phasor relation is VL+ VR+ VC= V

VL, and VC are along  same lines but in opposite directions, and phasor (VL+ VC) has magnitude vCm–vLm

With reference to the below image and using Pythagoras theorem, we can say that 

vm2=vRm2+(vCm–vLm)2 which implies

vm2=(ImR)2+(ImXC–ImXL)2 or

vm2= Im2(R2+(XC–XL)2)

Therefore, we can say that

Im= vm(R2+(XC–XL)2)

We introduce impedance Z such that

Z = R2+(XC–XL)2)

The Class 12 physics Chapter 7 notes have several problems based on impedance and application of AC voltage on the LCR circuit.  

Impedance diagram:-

It is defined as a right angle triangle with impedance (Z) as hypotenuse, base as R and perpendicular (height) as (XL–Xc)2.

is the phase angle between V and VR, V is the source voltage, and I is the current passing through the circuit. 

I is parallel to VR. 

tan = vCm–vLmvRm

Therefore, tan = XC–XLR

Here we have two cases, 

Case 1: XC>XL

The difference (XC–XL) is positive. Hence phase difference () will also be positive. XC will be more than XL. Hence, we can say that to Current (I) leads to Voltage (V) in the capacitive circuit. 

Case 2: XL>XC

The difference (XC–XL) is negative. Hence phase difference () will also be negative. XL will be more than XC. Hence, we can say that Voltage(V) leads Current (I) in the inductive circuit. 

Considering these two cases, students must practice many questions included in the Class 12 Physics Chapter 7 notes. 

Phasor Diagram for LCR Circuit:

The circuit elements Resistor, Inductor and Capacitor, are connected in series. 

I = Im sin (ωt+) is current, and V = Vm sin (ωt) is the voltage in the circuit. 

VL, VR, VC, and V are voltages in the circuit across the elements inductor, resistor, capacitance and source, respectively. And VL= ImXL, VR= ImR, VC= ImXC and V= Im

Since VR and Current are in phase. Current I lag behind VL, and VC lags behind IL

The Phasor diagram is represented as follows:

VL and VC are in opposite directions in the same line. 

From the Phasor diagram, we get that VL+ VR+ VC= V

Using Pythagorus theorem, we have

Z = R2+(XC–XL)2)

Image Name: Relation between Phasors 

[fig (a) depicts the relation between phasors VL, VR, VC and I. ( V-I relation) and fig (b) depicts the relation between phasors VL, VR, and (VL+VC) ]

Resonance: 

This phenomenon is an extraordinary characteristic of the LCR-circuit. Resonance is the tendency of a system to oscillate at greater amplitudes for some frequencies than others. These systems  have a common tendency to oscillate at a certain frequency. This frequency is said to be the natural frequency of the system. Consider the  following examples: 

1. A pendulum: 

The oscillations of the pendulum take place at its natural frequency. However, if a push is given, the amplitude increases. The frequency with which the pendulum now oscillates is known as the resonance frequency. 

1. Swing:

A child swings with the natural frequency of the swing. However, when someone applies a slight force or pushes the swing from behind at the same initial frequency, then the amplitude increases. This frequency is said to be the resonant frequency. 

Resonance of LCR circuit connected in series: 

We know that the amplitude is maximum at the resonant frequency. 

Therefore in the LCR circuit, the current amplitude is

im=vmZ = vmR2+(XC–XL)2), where XC= 1ωC and XL= L 

If the value of varies then at o, we can say that XC=XL and impedance Z =R

Therefore, XC=XL or 1ωC = L 

o=1LC is the value for resonant frequency. 

From the graph, the value of im increases with . It reaches a point o(maximum value) and then decreases again. 

It should be noted that resonance is exhibited only if an inductor with inductance L and Capacitor with capacitance C are present in a circuit. Then, the voltage across L and C can cancel each other, and current amplitude becomes the total source voltage passing across R. This means resonance cannot be obtained in a circuit with RL or RC. 

Application of Resonant Circuit: 

The resonant circuits have a wide range of applications. It is used in the tuning mechanism of a television or radio. The antenna of the radio picks up the signals from different broadcasting stations. The signals accepted by the antenna act as a source in the tuning circuit. The circuit is driven at many frequencies. To listen to a certain radio station, tuning is required. 

In tuning, the capacitance of a capacitor is varied in the circuit so that the resonant frequency becomes approximately equal to that of the actual frequency of the radio signal received. When this takes place, the amplitude of current with frequency of a particular radio station becomes maximum in the tuning circuit. Therefore we will be able to hear the sound. 

However, when the amplitude is minimum, the sound is not audible. When the amplitude is close to the peak value, we can hear the sound, but clarity won’t be there. . 

The sharpness of resonance:

The current amplitude in a resonant circuit is im=vmZ = vmR2+(XC–XL)2). It is maximum when o=1LC and the peak value is immax= vmR. 

Choose two values 1 and 2, such that 1 is greater than o and 2 is smaller than o.

Therefore, if is the difference, then we have

1 =o +

2 =o –

And 1 – 2 = 2 gives the bandwidth of the resonant circuit. 

Therefore, the sharpness of resonance is given as o 2 

We can say that, sharpness is inversely proportional to . Hence, smaller the value of , sharper is the resonance. 

o 2 =o LR , where o LR = Q (quality factor)

So, o Q= 2

If the value of Q is small, then 2 will increase. Thus, resonance will be sharper. 

If the sharpness of resonance is less, the maximum current is less, the circuit is close for larger of frequency and thus, tuning will not be good. Therefore, if the quality factor (Q) is large, or R is low, or L is large, then the circuit is selective. 

POWER ASSOCIATED WITH AC CIRCUIT

We know that voltage V = Vm sin (ωt) and I = Im sin (ωt+), where Im = VmZ and = tan-1(XC–XLR)

Therefore, instantaneous power (p) is given as

p = VI = Im Vm sin (ωt) sin (ωt+) 

Using 2 sinA sinB = cos(A-B) cos(A+B)

p = Im Vm [cos – cos(2ωt+)]

The average power for a complete cycle is the average of two terms on RHS. The second term is independent of time. Therefore, the average is zero. 

So we can say that 

P = VI cos or P = I2Zcos

The average power depends on voltage, current and cosine of the phase angle (). Therefore, cos is known as the power factor. 

Case 1: Circuit with only resistor

In this case, we consider a resistive circuit. Since = 0, therefore cos =1. Maximum power will be dissipated.

Case 2: Circuit with only inductor

In this case, we consider an inductive circuit. Since = 2, therefore cos =0. The voltage leads the current by 2. So, P will be zero. No power is dissipated. Here, the current is also called the wattless current. 

Case 3: Circuit with only capacitor

In this case, we consider a capacitive  circuit. Since = 2, therefore cos =0. The current is ahead of voltage by 2. so, P will be zero. No power is dissipated.

Case 4: Circuit with LCR

In an LCR circuit, = tan-1(XC–XLR). In this case, may be non-zero in RL, RC or  RLC  circuits. Therefore, power will be dissipated only by the  resistor. 

Case 5: At resonance in LCR circuit

Since, XC=XL, therefore = 0. Since, cos =1 and so, maximum power will be dissipated.

These concepts from the Class 12 Physics chapter 7 notes are very important. Students are advised to focus on this concept so that they can tackle any question from here  easily.   

LC OSCILLATIONS

LC circuit includes an inductor with inductance L and Capacitor with capacitance C. The energy which is taken from the cell passes through the capacitor and oscillates between L and C. These oscillations are known as LC oscillations. 

When AC voltage is applied to L, it charges and then discharges again repetitively. Once it is fully charged, it starts discharging and is transferred to the inductor. This leads to a change in current, accompanied by a change in the magnetic flux of the inductor. Therefore, EMF is induced in the inductor that tries to oppose the passing current. 

Consider a capacitor carrying charge qm at time t=0. It is connected to an inductor. As soon as the circuit is complete, the charge decreases and gives rise to current (i) in the circuit. So, i= – dqdt

We know that didt is positive. Therefore, the induced EMF in the capacitor has polarity vb < va.

Using Kirchoffs law, qC–didt= 0. 

Substituting value of current i as – dqdt, we get

d2qdt2–1LCq = 0

Therefore, o=1LC is a sin function and varies with time as q= qmcos(ot+), where qm is the peak value  of q. 

At t=0, q = qm. So, cos = 1 or = 0.

So we get, q= qmcos(ot)

The Electric and magnetic field energy in the circuit will keep oscillating.

The total energy of the LC circuit is U = UE= qm22C.

Total magnetic energy is UB= Lim22.

NOTE:

1. Every inductor has resistance. The effect of this resistance introduces a damping effect on the charge (q) and current (i) in the circuit, and the oscillations will eventually stop. 
2. When the resistance is zero, the total energy of the system does not remain constant. It radiates from the system in the form of electromagnetic waves. The transmitters of radio and TV depend on this radiation.

The below table lists the equivalence between Forced oscillations and Driven LCR circuit

This comparison of forced oscillations and Driven LCR circuits helps students to remember the concepts better so that they can solve all questions included in the Class 12 Physics Chapter 7 notes. 

TRANSFORMERS

Transformers are devices used to change or transform the values of alternating voltage. The power input is equal to the power output. 

Alternating Transformer

The transformers work on the mutual induction principle. Consider two inductors. Let current (I) flow through the first coil. There will be a change in the current as the magnetic flux changes. Therefore the magnetic flux in the second coil changes. Hence, EMF is induced in the coil. 

Construction: 

A transformer includes a primary coil with ‘NP’ number of turns over a soft iron core. It is the input end. The secondary coil also has ‘NS’ number of turns of a wire. It is the end where we receive the output. Soft iron core is an extremely thin piece. It covers the minimum possible area. Therefore the energy lost will be very less. A permanent magnet cannot be used in transformers  as the loss of energy will be huge. 

Working:

The input voltage (AC source) is applied through the primary coil. Therefore, in the primary coil, the alternating current is produced. That gives rise to magnetic flux in the coil. And thus, EMF is induced.

There are 2 EMFs produced. The first is self-induction, and the other is mutual induction. 

There is self-induced EMF in the primary coil as the magnetic flux changes in the primary coil. Therefore, magnetic flux will change in the secondary coil. Therefore, EMF is induced in the secondary coil. 

There is mutual induction in the secondary coil. 

The EMF induced in the primary coil is given as ep= –Np(ddt), where ddt is the rate of change of magnetic flux and Np is the number of turns in the primary coil. 

The EMF induced in secondary coil is given as es= –Ns(ddt), where ddt is the rate of change of magnetic flux and Ns is the number of turns in the secondary coil. 

Assume the resistance is 0 in both the coils. Therefore, ep= Vp. We can say that,

Vp= –Np(ddt) and es=Vs= –Ns(ddt)

Divide both the equations, we get

VpVS= NpNS

The above relation is obtained assuming that;

1. The resistance of the primary coil and current is very small.
2. The magnetic flux links the primary and the secondary coil is the same. Very less flux escapes into the surroundings from the core. 
3. The current passing through the secondary coil is small. 

Hence, VS=NsNpVp

We know that, 

Powerinput=Poweroutput

So, IpVp=IsVs

Since 100% energy is never obtained, some are always lost. Therefore, the efficiency of a well-designed transformer is 95%. 

NsNp=VsVp=IpIs

Current (I) and voltage (V) oscillate with the same frequency as that of the AC source. (I) and (V) both oscillate with the same frequency as the ac source,

Since, VS=NsNpVp, therefore Ip=NsNpIs

The topic of transformers is very important. The Class 12 Physics Chapter 7 notes include information about different types of transformers and their applications. 

TYPES OF TRANSFORMERS:

1. Step-up Transformer

The step-up transformer helps to amplify the voltage. The output voltage is higher than the input voltage, i.e., Vs>Vp only when Ns<Np and Ip<Is. 

We know that VS=NsNpVp. 

Therefore, the output voltage of the transformer is high, and the current is less. 

The Step-up transformers are used in power stations to supply the power to houses. 

1. Step-down Transformer: 

The output voltage is higher than the input voltage, i.e., Vs<Vp only when Ns<Np. 

Since, Ip<Is therefore Powerinput= Poweroutput. 

We know that VS=NsNpVp. 

Therefore, the output voltage of the transformer is low, and the current is high. 

The Step-down transformers are used in welding. 

Applications of Transformers

• Transmission of power: Consider the main power station. From this station, power is sent to sub-area power stations and thus to different houses. At this power station, a step-up transformer amplifies the voltage, and the current is reduced. When current is reduced, heat also reduces. The power loss minimises till it reaches sub-area stations. A step-down transformer is used at this substation. It reduces the voltage and supplies to houses. The power loss will  be very less as the distance between houses, and the substation is less. 
• Transformers help to regulate voltage. Many appliances regulate the voltage using voltage stabilisers so that no damage is done to the electronic devices whenever there is voltage fluctuation. 

Small energy losses occur in the actual transformers

1. Flux leakage:

The air gaps between the primary and secondary coils lead to the change in magnetic flux associated with the primary coil. Therefore, it is not completely transferred to the other coil. To reduce this wastage, the secondary coil is wounded over the primary coil. Considering toroidal transformer cores, the secondary coil is wounded above it. This leads to no air gap between the two coils. 

1. Resistance of windings:-

The wire which is used for windings has resistance. Therefore, heat is produced, which leads to a loss of energy in the wire. Since the area is directly proportional to resistance, if we increase the cross-sectional area of the wire, the value of resistance will also increase by a considerable amount. Using thick wires in the windings of the coils will help to reduce the resistance. The amount of heat lost because of the wires will reduce and become minimal. 

1. Eddy currents:

Due to the magnetic flux induced, the soft iron core heats up and eddy currents are developed in the iron core. We can say that, due to eddy currents, the core is heated which might tend to harm the transformer core. To prevent this harm, a laminated core can be utilised in place of the soft iron core. Because of the lamination (insulated covering), heating effects will not be produced by eddy current. 

1. Hysteresis:

The process of magnetisation of the material of the core is accompanied by energy loss. It is important that the chosen materials must have the lowest value of hysteresis, which is why the soft iron core is highly used in place of the permanent magnets.

Important definitions in the Class 12 Physics Chapter 7 notes

1. Alternating Current: It is the current which is different in both magnitude and direction. It changes alternatively and periodically. I = Im sinωt, where Im is the maximum or peak value of AC
2. RMS values of AC: It is a value which would generate the same amount of heat (sent by steady current) over a complete cycle in the given resistors in the same time t.
3. Average value of AC: It is the value which generates the same amount of charge in a half cycle throughout the circuit which is sent by a steady current. The average value of AC in a complete cycle is zero.
4.  Alternating EMF: It is said to be the value of EMF which varies in both magnitude and direction periodically and alternatively. It is given as V = Vm sinωt.
5. Inductive Reactance: It is denoted by XL. It is defined as the opposing nature of an inductor to the current passing through a circuit. XL=ωL
6. Capacitive Reactance: It is denoted by XC. It is defined as the opposing nature of a capacitor to the current passing through a circuit. XC= 1ωC
7. Power: The EMF and current in an AC circuit changes continuously with time. Therefore, the average power over a complete cycle is calculated as P = VI cos .
8. Average power consumption:   In a pure inductive or capacitive circuit, the average power consumption is zero as the phase difference is zero. 
9. Wattless current: When the average power consumption in a series AC circuit is zero, then the current passing through the circuit is known as the wattless current or  idle current. 
10. Step-up transformer: In this type of transformer, the input voltage is lower than the output voltage, i.e., Vs>Vp only when Ns<Np and Ip<Is. 
11. Step-down Transformer: In this type of transformer, the input voltage is lower than the output voltage, i.e., Vs<Vp only when Ns<Np. Since Ip<Is therefore Powerinput= Poweroutput. 

Important formulas in the Class 12 Physics Chapter 7 notes

1. AC Voltage- V = Vm sinωt
2. AC Current- I = Im sinωt
3. Capacitive  Reactance- XC= 1ωC
4. Inductive Reactance- XL=ωL
5. RMS Voltage- Vrms= V02
6. RMS Current- Irms= I02
7. Phase angle- tan = XC–XLR
8. Ohm’s law- Io= V0Z
9. Power Average- P = VI cos or P = I2Zcos
10. Resonant frequency- o=1LC 

Summary Of Chapter 7 Physics Class 12 Notes

According to the Class 12 Physics Chapter 7 notes, alternating current is the current which changes continuously in magnitude and polarity with time. This time interval remains fixed. It is described as an electric current which reverses the direction. In the CBSE revision notes, students will get a description of all concepts for quick revision.

Every definition in the Class 12 Physics Chapter 7 notes is explained in detail in an easy language and lucid manner. It also has several graphs and illustrations wherever necessary. These graphs represent different types of current and explain the difference between them. Students will be able to memorise these concepts very easily. Referring to  Class 12 Physics Chapter 7 notes will help students to eliminate all doubts and cover the entire chapter quickly and efficiently. 

Students will learn about the formation of AC current, the devices used and the generation of the current. A detailed explanation of Faraday’s principle induces a current with the help of mechanical energy is given. Topics such as uses and advantages, waveform and frequency of AC have been included in the Class 12 Physics Chapter 7 notes. Students will understand how to determine the average value of AC, root mean square values using the formulas included in this chapter. In proceeding with the chapter, concepts of phasor diagrams, AC circuit designs, transformers, and LC oscillations are explained in a step-wise manner in the Class 12 Physics Chapter 7 notes. This chapter is very crucial as it lays the foundation for students  interested in pursuing  electrical engineering as their career. 

Class 12 Physics Chapter 7 notes: Exercises &  Solutions

The Class 12 Physics Chapter 7 notes are essential to gain in-depth knowledge of Alternating Currents. Students will have to solve various CBSE extra questions based on the power associated with AC circuits, Alternating Voltage and current in resistive, inductive, capacitive  or LCR circuits, etc. With the help of Class 12 Physics Chapter 7 notes, students can ensure the consistent practice of all objective and subjective questions. The key points, formulas and summary included in the notes will help students to revise quickly and also retain    their information for a longer period of  time. 

Class 12 Physics Chapter 7 notes provide a helping hand to students for their Board exam and competitive exams preparation. The notes contain logical solutions to all the important questions and exercise problems  included in the CBSE Syllabus. Each solution is detailed and well explained in a step-wise manner in the Class 12 Physics Chapter 7 notes. Studying with the help of these notes will help students to clear their doubts and boost their confidence to get  good marks.  

The Extramarks Class 12 Physics Chapter 7 notes aims to strengthen the foundation by providing clear understanding of concepts and core knowledge. It provides a revision of formulas, definitions, derivations and key points in a descriptive manner. In case of any  queries, with the help of the Class 12 Physics Chapter 7 notes, students can face every complication with ease and continue with their preparation confidently. . 

Students are advised to refer to different academic materials, and study notes like the Class 12 Physics Chapter 7 notes provided on the Extramarks web portal to prepare for the Class 12 boards as well as other national level examinations such as the JEE Main, JEE Advanced, NEET, etc.

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Key Features of Class 12 Physics Chapter 7 notes

 Extramarks Class 12 Physics Chapter 7 notes: Key Features 

• The Class 12 Physics Chapter 7 notes are prepared by subject matter experts at Extramarks. The team has analysed various CBSE sample papers to make sure that the information provided is accurate. 
• The information provided in the Class 12 Physics Chapter 7 notes follows all the latest rules and  regulations issued by the CBSE board. It  is based on the latest NCERT Syllabus of 2021-22. 
• Students can easily access the Class 12 Physics Chapter 7 notes from any electronic device such as a mobile, laptop, computer, etc., just download the mobile version or visit the Extramarks website to access all information related to  CBSE exams.
• The Class 12 Physics Chapter 7 notes ensure regular study and practice of all problems and concepts included in the chapter. Also, students will gain information about the weightage and marking system of Class 12 boards. 
• Studying with the help of Class 12 Physics Chapter 7 notes helps to develop analytical and problem-solving skills. Taking various time-bound tests also is important to develop time-management skills. It will boost their confidence level  and reduce stress and anxiety to a great extent.