Different types of circuit

-

1. AC circuit containing resistance only



We have

E = Eosinwt        ..... (1) 


If I be the current flows through R
Then, acc to ohm's law

E = IR

I = E/R

Now from eq. (1) 

I = Eo sinwt/R

I = (Eo/R) sinwt

I = Io sinwt     ..... (2) 

Where Io = Eo/R

Thus, it is clear that E and I are in same phase

Phaser diagram


2. AC circuit containing inductor only



We have 
E = Eosinwt   ... (1) 
As I flows through inductor then acc. To faraday law in induced EMF -LdI/dt produced in it which opposes the external EMF E of source. 

Net EMF = E - LdI/dt

As there is no resistance connected through the circuit

Then acc. To ohm's law

Net EMF = IR
But R = 0
Such that net EMF = 0
Then, 

E - LdI/dt = 0

E = LdI/dt

From eq. (1) 

Eosinwt = LdI/dt

dI = Eosinwtdt/L

dI = Eo/L [sinwt dt]

On integrating both sides we get

I = Eo/L [ -coswt/w ]

I = - (Eo/Lw) coswt

I = - Io coswt

I = - Io sin(π/2 - wt) 

I = Io sin(wt - π/2 ) 

On comparing equation (1) and (2) 

It is clear that EMF leads current by angle π/2.

Phaser diagram



              

                            Note(1) 


We have 

Io = Eo/wL

Wkt
Acc to ohm's law

Current = EMF/resistance

Here, resistance = wL = xL = inductive resistance

XL = wL

It is the effective resistance offer by circuit containing inductor only. 

                             Note(2) 


We have xL = wL
But 
W = 2π/T = 2π(1/T) = 2πf

xL = 2πfL

For DC
f = 0
xL = 2π × 0 × L
xL = 0
It means, DC flows easily through  inductor. 

For AC
xL is directly proportional to f

It means, AC does not flows through inductor, it offers high resistance for it. 


3. AC circuit containing capacitor only




Consider a 2 capacitor having Capacitance C and charge Q connected through EMF E


We have E = Eosinwt    .... (1) 
WKT
Q = CV(V = E) 
Q = CE
Also WKT
I = dQ/dt

I = d(CE) /dt
From eq. (1) 

I = d(CEosinwt) /dt

I = CEo × d(sinwt) /dt

I = CEO [coswt] × w

I = Io coswt     because Io = CEow

I = Io sin(π/2 + wt) 

I = Io sin(wt + π/2)    ...... (2) 

On comp. Eq. (1) and (2) 
It is clear that 

I leads E by an angle π/2

Phaser diagram


                    Note(1) 

We have
Io = EoCw
Or
Io = Eo/(1/Cw) 

WKT acc to ohm's law
 Current = EMF/resistance

Resistance = 1/wC = capacitance resistance = xC

So it means, 


      xC = 1/wC

                           Note(2) 


We have 
       xC =1/wC

     But w = 2πf

For DC
As       xC is inversely proportional to frequency

xC = infinity
Because for DC   f = 0

So, it means DC does not flows through capacitor. 

And

For AC  
As we know xC is inversely proportional to f 

So it means AC flows easily through capacitor. 

4. AC circuit containing L&R



Phaser diagram



Now, the resultant of Er and El

Where, Er = EMF on resistance R 
And El = EMF on inductor L

E × E = El × El + Er × Er
E × E = (IR) × (IR) + (I × xL) × (I × xL)

E = I (R × R + xL × xL)  where bracket has power of 1/2

I = E/Z

Where, Z = impedance = (R×R + xL×xL) where bracket has power of  1/2

Phase angle

Tan(phie) = El/Er = I xL/R

Also cos(phie) = Er/E

Cos(phie) = IR/I(R×R + xL×xL) where bracket has power of 1/2

Cos(phie) = R/(R×R + xL×xL) where bracket has power of 1/2

Cos(phie) = R/Z         {power factor}


5. AC circuit containing R and C


Phaser diagram




Now the resultant of Er and Ec
Where Er is EMF on resistance and Ec is EMF on capacitor

E×E = Ec×Ec + Er×Er
E×E = (Ir×Ir + I xC × IxC) 

E = I (R×R + xC × xC) where bracket is of power of 1/2


I = E/Z

Where, Z = impedance = (R×R + xC×xC) where bracket is of power of 1/2.

Phase angle


Tan(phie) = Ec/Er = IxC/Ir = xC/R

Also cos(phie) = Er/E

Cos(phie) = IR/I(R×R + xC×xC) 

Cos(phie) = R/(R×R + xC×xC) 

Cos(phie) = R/Z                        {power factor}
Where, Z = (R×R+xC×xC) 


6.series LCR circuit




V = IR
Now, acc to ohm's law 
Er = IR
EL = IxL
Ec = IxC

Phaser diagram





Now the resultant of Er and (El-Ec)

E×E = Er×Er + (El-Ec) × (El-Ec) + 2(Er) (El-Ec) cosπ/2

E×E = Er×Er + (El-Ec) × (El-Ec) + 0

E×E = (IR) (IR) + [IxL - IxC] [IxL - IxC]

On solving we get

I = E/[R×R + (xL-xC) (xL-xC) ]   where denominator bracket has power of 1/2

I = E/Z

Where, 
         Z = [R×R + (xL - xC) (xL - xC) ]     where bracket has power of 1/2


Phase angle between E & Er


Tan(phie) = El-Ec/Er

Tan(phie) = (IxL - IxC)/ IxR

Tan(phie) = (xL - xC) / R

Also cos(phie) = Er/E

Cos(phie) = Ir/I[R×R + (xL-xC) (xL-xC) ] where denominator bracket has power of 1/2

So, 

Cos(phie) = R/[R×R + (xL-xC) (xL-xC) ]  = R/Z      {power factor}

Resonant condition for series LCR circuit


The resonant condition is said to be obtained when the current flows through the series LCR circuit becomes maximum, 
We have, 
I = E/Z = E/[R×R + (xL-xC) (xL-xC) ] where denominator bracket has power of 1/2

Current becomes maximum when xL = xC
Then, I = E/[R×R + (xL-xC) (xL-xC) ] where denominatorbracket has power of 1/2

I = E/(R×R) where denominator bracket has power of 1/2

I = E/R
So, I is maximum when R is minimum

Resonant frequency


WKT 
At resonant condition 
xL = xC
wL = 1/wC
w × w = 1/LC

wr = 1/(LC)  where denominator bracket has power of 1/2       .... (1) 

Where, wr  = angular speed at resonant condition
Also we know that

wr = 2πfr      (2) 
Where fr is resonant frequency
 
So from eq. (1) and (2) 

2πfr = 1/(LC)  where denominator bracket had power of 1/2

fr = 1/2π(LC)          where denominator bracket has power of 1/2



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