Alternating Current (AC): Class 12 Notes, Definition, Working Principle & Diagram

Candidates can check here the list of topics that are covered in the Alternating Current chapter. The detailed analysis for every Alternating Current chapter topic is also provided on this page.

Alternating Current Topics
Transformer	Kirchhoff's Law	Ohm’s Law
Coulomb’s Law	Electric Dipole	SI Unit of Conductivity
Milliampere to Ampere	Electron Drift Velocity	Unit of Molar Conductivity
Difference of AC and DC	Superposition Principle	Heating Effect of Electric Current
Electric Potential due to Point Charge	Half Wave Rectifier Diagram	Series and Parallel
Unit of Specific Conductivity	Unit of Equivalent Conductivity	Electric Charge
Gauss Law	Electric Flux	Application of Gauss Law
Potentiometer	Temperature Dependence of Resistivity	Combination of Resistance – Series and Parallel

“A rotating vector that represents a varying physical quantity is called a Phasor.” The length of the vector represents the magnitude and its projection on a fixed axis gives the instantaneous value of the quantity To study the alternating current, Alternating voltage and current are considered as rotating phasors.

“Phasor diagram are diagram representing alternating current and voltage of same frequency as vectors or phasors with the phase angle between them and rotating about the origin.”

The vertical components of phasors V and I represent the sinusoidally varying quantities v and i. The angle between the V and I phasor is known as phasor angle or simply Phase(∅)

Let a resistor of resistance R is connected to a source of an alternating emf is

𝜺 = 𝜺𝒐 𝒔𝒊𝒏 𝝎𝒕 − − − −(𝟏)

Let I be the current in the circuit at any instant of time t , then potential across resistor is given by ohm’s law,

𝐼 =𝜀/𝑅 = 𝜀𝑜 sin 𝜔𝑡/𝑅

𝑰 = 𝑰𝒐 𝒔𝒊𝒏 𝝎𝒕 − − − −(𝟐)                 (∵    𝑰𝒐= 𝜺𝒐 /𝑹 )

Comparing equation (1) and (2) , Current I and the voltage(emf) 𝜺 both are in the same phase. This means that both I and 𝜺 attains the maximum, minimum and zero at the same time. The phasor diagram for purely resistive circuit is shown in below diagram.

The phase angle between the current I and 𝜀 is zero. i.e. (∅ = 0^𝑜)

Let an Inductor, L is connected to source of alternating current , 𝜀 as shown in fig(a). The varying emf is given by,

𝜺 = 𝜺𝒐 𝒔𝒊𝒏 𝝎𝒕 − − − −(𝟏)

The circuit is now purely Inductive a.c. circuit. Let the current I flows through the inductor, then induced emf (reverse direction) is generated due to EMI which oppose the applied emf.

𝜀 = 𝐿 𝑑𝐼/𝑑𝑡

𝜀𝑜 𝑠𝑖𝑛 𝜔𝑡 = 𝐿 𝑑𝐼/𝑑𝑡          𝑓𝑟𝑜𝑚 𝑒𝑞𝑢𝑎𝑡𝑖𝑜n(1)

𝑑𝐼 = 𝜀𝑜/ 𝐿 𝑠𝑖𝑛 𝜔𝑡 . 𝑑𝑡

Integrating above equation we get

∫ 𝑑𝐼 = ∫ 𝜀𝑜 / 𝐿 𝑠𝑖𝑛 𝜔𝑡 . 𝑑𝑡

𝐼 = −  𝑜  𝑐𝑜𝑠 𝜔𝑡 ⟹ 𝐼 = −𝐼 cos 𝜔𝑡

𝜔𝐿

where, 𝐼𝑜   

                =  𝗌𝑜 

is the peak value of the current.

𝐼 = −𝐼𝑜 𝑠𝑖𝑛 ( -𝜋/2− 𝜔𝑡)

∵ [cos 𝜃 = 𝑠𝑖𝑛 (-𝜋/2 𝜃)]

𝑰 = 𝑰𝒐 𝒔𝒊𝒏 (𝝎𝒕 −𝝅/𝟐) − − − −(𝟐)

Comparing equation (1) and (2), current I lags behind the emf 𝜺 by 𝟗𝟎^𝒐. It means the phase

angle between I and 𝜺 𝑖𝑠  𝝅/𝟐 . i.e. (∅ = 90^𝑜) as shown in phase diagram (b).

It is the effective resistance offered by the inductor to the flow of

A.C. It has unit ohm(Ω).

We know   𝐼𝑜 =  𝗌𝑜/𝜔𝐿

and  𝐼𝑜= 𝗌𝑜/𝑅

Comparing above equation we write, 𝑅 = 𝑿𝑳 = 𝜔𝐿

Where, f is the frequency of A.C. supply.

 

• For C. ⟹ 𝑿𝑳∝ 𝑓.
• For C. ⟹ f=0 hence 𝑿𝑳= 0
• Thus an inductor allows D.C. through it Easily, but opposes the flow of A.C.

Let a capacitor, C, be connected to an alternating emf source. The capacitor get charged during half cycle and discharges during the next half cycle. As result there is a continuous current in the circuit as shown in fig(a) below.

We know ,

𝜺 = 𝜺𝒐 𝒔𝒊𝒏 𝝎𝒕 − − − −(𝟏)

At any instant,

Potential across capacitor(V) = Applied emf(𝜺)

𝑉/𝐶 = 𝑄 ⟹ 𝑄 = 𝐶𝑉

𝑄 = 𝐶𝜀                (∵ 𝐶 =  ) , 𝑓𝑟𝑜𝑚 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛(1)

𝑄 = 𝐶𝜀𝑜 𝑠𝑖𝑛 𝜔𝑡

By definition of the current at any instant of time ‘t’ is 𝐼 = 𝑑𝑄/𝑑𝑡

𝐼 =𝑑(𝐶𝜀𝑜𝑠𝑖𝑛 𝜔𝑡)/𝑑𝑡  ⟹ 𝜔𝐶𝜀𝑜𝐶𝑜𝑠𝜔𝑡

𝜀_𝑜𝐶𝑜𝑠𝜔𝑡 𝐼 =1/𝜔𝐶 = 𝐼_𝑜𝐶𝑜𝑠𝜔𝑡    where,      _𝑰𝒐= 𝜺_{𝒐/ (𝟏⁄  𝝎𝑪 ) = 𝑝𝑒𝑎𝑘 𝑐𝑢𝑟𝑟𝑒𝑛𝑡}

𝑰 = 𝑰𝒐𝒔𝒊𝒏 (𝝎𝒕 + 𝝅/𝟐) − − − (𝟐)

Comparing equation (1) and (2) we can say that current leads the voltage(emf) by angle  𝟗𝟎^𝒐

shown in the phasor diagram below(b).Also phase angle is ∅ = 𝝅/2 = 𝟗𝟎^𝒐

It is the effective resistance offered by the capacitor to the flow of A.C. It has unit ohm(Ω).

We know  𝑰𝒐=      𝜺𝒐 (𝟏⁄𝝎𝑪)

and  𝐼𝑜 = 𝗌𝑜/R

Comparing above equation we write, 𝑅 = 𝑿𝑪 = 1⁄𝜔𝐶

Where, f is the frequency of A.C. supply.

• For A.C. ⟹ 𝑿𝑪 ∝ 1/𝑓
• For D.C. ⟹ f = 0 hence 𝑿𝑪= ∞

Thus a capacitor allows A.C. through it easily but blocks the flow of DC rather it sores all of its energy. through it.

Suppose a resistance R, inductor L, and capacitor C are connected in series to an alternating source of emf 𝜀 given by,

𝜺 = 𝜺𝒐 𝒔𝒊𝒏 𝝎𝒕 − − − −(𝟏)

Phasor diagram: For LCR circuit

(b) Impedance(Z) triangle

From circuit diagram, if I be the current in the circuit then applying ohm’s law to each component,

𝑉𝑅 = 𝐼𝑜𝑅                𝑉𝐿 = 𝐼𝑜𝑋𝐿               𝑉𝐿 = 𝐼𝑜𝑋𝐶 − − − − − (2)

𝑉𝐿   𝑎𝑛𝑑 𝑉𝐶 are in opposite direction, their resultant is ( 𝐿 − 𝑉𝐶 ).Hence the resultant of (𝑉𝐿 − 𝑉 ) and 𝑉𝑅 is equal to the applied emf (𝜺).

(𝜺𝒐)^𝟐 = (𝑽𝑹) + ( 𝑉𝐿 − 𝑉𝐶 )^𝟐                                    by Pythagoras thm.

From equation (1) and (2

𝜺𝒐 = √(𝑽𝑹)^𝟐 + ( 𝑉𝐿 − 𝑉

𝜺𝒐 = 𝑰𝒐√(𝑹)^𝟐 + ( 𝑋𝐿 − 𝑋𝐶 )

Hence

𝑰𝒐 =                𝜺𝒐                       ⟹ 𝜺𝒐  /z

√(𝑹)𝟐 + ( 𝑋𝐿 − 𝑋𝐶 )𝟐

𝒁 = √(𝑹)^𝟐 + ( 𝑿𝑳 − 𝑿𝑪)^𝟐

(h2) Impedance(Z):It is an effective resistance offered by LCR-circuit to the flow of an A.C. current through it. It has unit ohm(Ω).

𝒁 = √(𝑹)^𝟐 + ( 𝑿𝑳 − 𝑿𝑪)^𝟐 = √(𝑹)^𝟐 + ( 𝝎𝑳 − 𝟏/𝝎𝑪 )^𝟐

Special cases

1)  When 𝑿𝑳> 𝑿𝑪 then emf is ahead of current by phase angle 𝟗𝟎^𝒐 which is given by,

𝐭𝐚𝐧 ∅ = 𝑿𝑳 − 𝑿𝑪/𝒁  𝒐𝒓  𝐜𝐨𝐬 ∅ = 𝑹 / 𝒁

𝑇ℎ𝑒 𝐿𝐶𝑅 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑖𝑠 𝑝𝑢𝑟𝑒𝑙𝑦 𝐼𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑒

2) When 𝑿𝑳< 𝑿𝑪 then current is ahead of emf by phase angle 𝟗𝟎^𝒐 which is given by,

𝐭𝐚𝐧 ∅ = 𝑿𝑪 − 𝑿𝑳/𝒁  𝒐𝒓 𝐜𝐨𝐬 ∅ = 𝑹 / 𝒁

𝑇ℎ𝑒 𝐿𝐶𝑅 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑖𝑠 𝑝𝑢𝑟𝑒𝑙𝑦 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑖𝑣𝑒

3) When 𝑿𝑳= 𝑿𝑪 then current and emf both are phase angle e. Ø=0

𝐼𝑜 = 𝜀𝑜/𝑍     𝒐𝒓    𝐼𝑟𝑚𝑠 = 𝜀𝑟𝑚𝑠/𝑍

𝑇ℎ𝑒 𝐿𝐶𝑅 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑖𝑠 𝑝𝑢𝑟𝑒𝑙𝑦 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑖𝑣𝑒

A LCR circuit is said to be in the resonance condition when the current through it has a maximum value.

Resonant Frequency(fr): The frequency at which the current through LCR circuit is maximum. At resonant frequency the impedance(Z) is minimum. The resonance occurs when

𝑋𝐿 = 𝑋𝐶

2𝜋𝑓𝐿 = 1/2𝜋𝑓𝐶

𝑓² = 1/4𝜋²𝐿𝐶   ℎ𝑒𝑟𝑒     𝑓𝑟 = 𝑓

The resonant frequency ( $f_r$) is given by the formula:

$$f_r = \frac{1}{2\pi \sqrt{LC}}$$

Note: 1)At resonance current and voltage are in same phase i.e Ø=0. The circuit is purely resistive. The current value is given by ,  𝐼𝑜 = 𝗌o/𝑅

2)At resonance the power factor is unity 𝐜𝐨𝐬 ∅ = 𝟏

 

 

The sharpness of the resonance depends on the resistance value. In simple terms, we can say the smaller the resistance the sharper the curve of the current's amplitude will be. If the resistance is small then resonance curve has sharp peak, and flat for the large resistance. The sharpness of resonance is measured by quantity called as Quality Factor(Q).

Quality Factor(Q):It is definedas the ratio of the angular resonant frequency to the band width of the circuit. It is given by,

It is a dimensionless quantity

The rate of the electrical energy consumed or produced by electric current in an electrical circuit is called as power.

The instantaneous power is given by,

𝑃𝑎𝑣 = 𝜀𝑟𝑚𝑠 𝐼𝑟𝑚𝑠 cos ∅  but cos ∅ = 𝑅/𝑍

𝑹

𝑷𝒂𝒗 = 𝜺𝒓𝒎𝒔𝑰𝒓𝒎𝒔. 𝒁

The power factor is defined as the ratio of the true power to the apparent power of an A.C. circuit. It is dimensionless quantity. It is given by

cos ∅ = 𝑅/𝑍 =                        𝑅                   

                                                       𝟏  ^𝟐

(𝑹) + ( 𝝎𝑳 −𝝎𝑪)

 

Special cases:

Purely resistive circuit: Here ∅ = 0 and cos ∅ = 1

𝑃𝑎𝑣 = 𝜀𝑟𝑚𝑠𝐼𝑟𝑚𝑠 × 1 = 𝜀𝑟𝑚𝑠 × 𝐼𝑟𝑚𝑠

𝑷𝒂𝒗 = (𝜺𝒓𝒎𝒔)𝟐/𝑹

Purely inductive circuit: Here ∅ = 𝜋and cos ∅ = 0

𝑷𝒂𝒗 = 𝟎

Purely capacitive circuit: Here ∅ = 𝜋and cos ∅ = 0

Ø  In series LCR circuit: 𝑷𝒂𝒗 = 𝟎

𝑃𝑎𝑣 = 𝜀𝑟𝑚𝑠𝐼𝑟𝑚𝑠 cos ∅

Where ∅ =

Power dissipated at resonance condition in LCR-circuit:

Here ∅ = 0 and cos ∅ = 1 also 𝑿𝑪 = 𝑿𝑳

𝑃𝑎𝑣 = 𝜀𝑟𝑚𝑠𝐼𝑟𝑚𝑠

Wattles Current: The current in the A.C. circuit is said to wattles if the power consumed in the circuit is ZERO.i.e. 𝑃𝑎𝑣 = 0 (1marks)

Numerical: A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mh, and C = 796 µf. Find

(a) the impedance of the circuit;

(b) the phase difference between the voltage across the source and the current;

(c) the power dissipated in the circuit; and

(d) the power factor

LC- OSCILLATIONS: When a charged capacitor is allowed to discharge through a inductor, an electrical oscillation of constant amplitude and frequency are produced. These oscillations are called as LC- oscillations.

Frequency of oscillation is given by,

𝒇 =𝟏/𝟐𝝅√𝑳𝑪

                                                        

Mechanical system	Electrical system
Mass, m	Inductance, L
Displacement , x	Charge, q
Velocity, 𝑣 = 𝑑𝑥/𝑑𝑡	Current, 𝑖 = 𝑑𝑞/𝑑𝑡
Force constant, k	Reciprocal capacitance, 1/𝐶
Mechanical energy 𝐸 = 1 𝑘𝑥2 /2+ 1 𝑚𝑣2 /2	Electromagnetic Energy 𝐸 =  𝑞2 / 2𝐶  +    1/2𝐿𝐼2
Frequency, 𝒇 =   𝟏 / 𝟐𝝅     √𝒌/m	Frequency, 𝒇 = 𝟏 / 𝟐𝝅√𝑳𝑪

LC oscillator are damped due to following reasons:

Means amplitude of oscillation decreases due to,

1) Every inductor has some resistance

2) total energy of the system would not remain constant

A transformer is a device that transfers electric energy from one alternating-current circuit to one or more other circuits, either increasing (stepping up) or reducing (stepping down) the voltage.

Transformers are used in many applications, including power generation, transmission, distribution, and electronic equipment. 

How transformers work

• Transformers use the principle of mutual induction to transfer energy between circuits. 
• A transformer consists of two coils, a primary winding and a secondary winding. 
• When the primary winding is connected to an alternating voltage source, an alternating magnetic flux is created around the winding. 
• This magnetic flux links with the secondary winding, inducing an EMF in it. 

Types of transformers 

• Step-up transformers: Increase voltage
• Step-down transformers: Decrease voltage
• Iron core transformers: Use laminated iron sheets for their magnetic properties
• Laminated core transformers: Use laminated sheets of iron and nickel to reduce energy losses

Applications of transformers: Power generation, transmission, and distribution, Lighting, Audio systems, Electronic equipment, Signal processing, Measurement, and Pulse transmission.

Candidates are provided here the Physics class 12 chapter 7 notes PDF that can be used as a resource to prepare for the board exams as well as competitive exams like JEE Mains and NEET. The Physics class 12 chapter 7 notes have been prepared by subject experts and include important alternating current formula and equations along with short notes that are easy to remember. Also check NCERT Solutions for Class 12 Physics Chapter 7 PDF here.

Download Alternating Current Physics class 12 chapter 7 notes PDF

Alternating current is one of most important chapter for JEE Main and NEET exams. Every year atleast one question is asked from this chapter which makes it must read chapter for NEET and JEE Mains preparation. The important topics from AC for JEE and NEET are

• Power Factor
• Impedance
• Resonance in AC Circuits
• Power calculations in AC Circuits
• Resistors, Capacitors and Inductance

Candidates can also check NEET and JEE Mains Previous year question papers to check what type of AC questions asked in the exam in previous years.

Alternating Current is the source for running all electrical appliances in our homes be it Television, Refrigerator, Air Conditioners, Washing Machine to name a few.

Alternating Current also have many practical applications in commercial spaces like running heavy machineries implemented for productions, lathe machines among others.

In healthcare sector, AC is used to run X-Ray, MRI, CT Scan and other pathology machines.

Alternating current is the power which runs trains and metro in cities. Electric vehicles are charged by Alternating Current.