We will cover the importance of AC current, give a brief introduction to the topic, and also provide refernces to useful content.

### Importance

AC current is in the syllabus for both JEE mains and JEE Advanced. In the last 5 years nearly 9 questions came in JEE mains. Counting the online exams as separate exams, we see that approximately 1 question comes from this topic comes every year in JEE mains. Similarly, in JEE Advanced, 6 questions were asked in the last 5 years.

### What is AC current or AC voltage?

Voltage and current from a source can change with time, like other physical quantities such as heat generated from source and temperature of a vessel.The reasons as to why most current generated is AC is also an interesting discussion, however we shall not be considering this here.

Here is an interesting video describing AC Current

So now how can we find the voltage at any point of time? One way is that we have a voltmeter attached to it and are noting down values. However this is not very convenient, especially if you have a large circuit. Instead, we try to represent it with an equation. Generally the voltage varies sinusodially with time. So we can write V=V_msinomegat, where V_m is the peak voltage and omega is the frequency of the voltage wave. This means that at time t=0 the volatge is zero, and at time t=pi/(2omega) the voltage will be maximum. The current flowing through a wire can also be represented in a similar way.

### RMS Value

In the previous section we talked about the current and volatge at any point of time being a function of the maximum current that follows through the circuit and time. Another term of importance in understanding AC circuits is the RMS voltage or current. Read more about this from the Nuffief Foundation

#### Resistor across alternating voltage:

Let us now try to understand the effect of applying an alternating voltage across a resistor and study its characteristics.
From Kirchoff's loop law, we get,

epsilon-iR=0=>i=epsilon/R=V_m/Rsinomega t=i_msinomega t, where i_m and V_m are the maximum value of current and voltage respectively.

To calculate the power dissapated in an electric circuit we us the formula P=i^2R where P is the power dissapated at an instant where current is i.

This is averaged over time period of the wave to obtain average power dissapated.

< P > = < i^2R > = R< i^2 > =R(int _0 ^T i_m^2sin^2omegatdt)/(int _0 ^Tdt)=1/2i_m^2R, T=(2pi)/omega

Hence average power dissapated barP = 1/2i_m^2R =(i_m/sqrt2)^2 R=i _(rms)^2R

and i _(rms) or I is defined as root mean square velocity and is sqrt((int _0 ^Ti^2dt)/(int _0 ^Tdt))

Similarly, we can define v _(rms) or V is defined as root mean square voltage and is sqrt((int _0 ^Tv^2dt)/(int _0 ^Tdt))=v_m^2/2 (for a sinosuidal profile)

We know that, v _m=i _mR=>v _m/sqrt2=i _m/sqrt2R=>V=IR

Hence average power loss is given by barP=I^2R=VI=V^2/R

Phasor

Dr Doc has explained a phasor very nicely in this video
A detailed explanation of phasors is available here, which you can probably read afterwards.

#### Capacitor across alternating voltage:

Now we see what happens when alternating voltage is applied across a capacitor.

Again from Kirchoff's loop law, we get,

epsilon-q/C=0=>q=Cepsilon=v_mCsinomegat
(we know i=(dq)/(dt)) =>i=d/(dt)(v_mCsinomegat)=v_momegaCcosomegat=v_m/(1/(omegaC))cos(omegat)=i_msin(omegat+(pi)/2)
i_m=v_m/(1/(omegaC)) where v_m is the peak value of voltage and 1/(omegaC) is called 'Capacitive reactance' denoted by X_C and units Ohm(Omega).

Phasor Diagrams

Dr Doc's explanation of phasor here

Instantaneous power

P=vi=(v_msinomegat)(i_mcosomegat)=v_mi_msinomegatcosomegat=1/2v_mi_msin2omegat

Now try the following yourself. The average power over a complete cycle (similar to what we did in the case of resistors).

< P > =< vi >

This should turnout to be 0.

#### Inductor across alternating voltage:

When alternating voltage is applied across an inductor its characteristics are as follows.
Applying Kirchoff's loop law, we get,

epsilon-L(di)/(dt)=0=>(di)/(dt)=epsilon/L=v_m/Lsinomegat =>di=v_m/L intsinomegatdt=-v_m/(omegaL)cosomegat=v_m/(omegaL)sin(omegat-(pi)/2)=i_msin(omegat-(pi)/2), i_m=v_m/(omegaL) where v_m is the peak value of voltage and omegaL is called 'Inductive reactance' denoted by X_L and units Ohm(Omega).

Phasor Diagrams

Doc Video

Instantaneous power

P=vi=v_msinomegat(-i_mcosomegat)
=-v_mi_msinomegatcosomegat=-1/2v_mi_msin2omegat

Now try to compute the average power in this case also.
< P > =< vi >

This should turnout to be 0. Thus, the average power supplied to an inductor over a complete cycle is zero.

#### RC, LR, LCR circuits

Now that we know the basics lets combine them.

You can see a video talking about RC circuits here another talking about LC circuits here. A detailed analysis of series LCR circuits is available at electronics-tutorials. Detailed analysis of parallel LCR circuits is available at electronics-tutorials.