B.Sc. Sem II Network theorems

1 | U V , S e m e s t e r - I I ( P A N a g p u r e )
UNIT: V
NETWORK THEOREMS
Kirchhoff’s Laws
The laws are:
Kirchhoff’s Current Law (KCL)/ Kirchhoff’s First Law/the Junction
Rule:
For a given junction or node in a circuit, the sum of the currents entering
equals the sum of the currents leaving. It supports the law of conservation
of charges. For example, in fig. the junction rule tells us
1 2 3 5 4 6I I I I I I    
The currents I1, I2, I3 and I5 are entering (flowing into) the junction, while the currents I4
and I6 are leaving (flowing out of) the junction.
Kirchhoff’s Voltage Law (KVL) Kirchhoff’s Second Law/the Loop Rule:
Around any closed loop in a circuit, the algebraic sum of the potential differences (voltages)
across all elements is zero. Or "The algebraic sum of all voltages in a loop must equal zero" It
supports the law of conservation of energy.
Sign convention:
While applying KVL for algebraic sum of the voltages across the elements, sign conventions to
be used are as follow-
i. The currents which are in the same direction in which the loop is traced (which can be
done either clockwise or anticlockwise) are treated as positive, while those in
opposite direction are treated as negative.
ii. The e.m.f. of a cell treated as negative, if it sends the current in the direction in which
loop is traced; otherwise the e.m.f. is treated as positive.
Consider loop ABCDA of figure 2. Applying KVL by proceeding clockwise, yields.
1 1 2 2 1 0I R I R V   
Consider loop BEFCB of figure 2. Applying KVL by proceeding clockwise, yields.
2 2 3 3 2 0I R I R V  
Consider loop ABEFCDA of figure 2. Applying KVL by proceeding clockwise, yields.
1 1 3 3 1 2 0I R I R V V    

Thevenin’s theorem
Statement: A linear network consisting of a number of voltage sources and resistances can be
replaced by an equivalent network having a single voltage source VTH (called Thevenin’s
voltage) and a single series resistance RTH (called Thevenin’s resistance).
Explanation:
Consider a network or a circuit as shown. Let E be the emf of the cell having its internal
resistance r = 0 and RL be the load resistance across AB.
To find VTH:
VTH is the open circuit voltage across AB when load resistance RL is removed.
The current I in the circuit is
Therefore
To find RTH:
RTH is the effective resistance across AB when load resistance RL is removed and voltage source
is replaced by its internal resistance (r=0).
Therefore,
[R1is parallel to R2and this combination is in series with R3]

If the cell has internal resistance r, then
Proof of Thevenin’s theorem:
Consider the network as shown below
The equivalent circuit is given by
The effective resistance of the network in (1) is R3and RL in series and this combination is
parallel to R2 which is in series with R1. Thus,
The current I in the circuit is
or
The current through the load resistance (I’) is found using branch current method.
Substituting for I from (2) in (3)
or

Thevenin’s voltage VTHand Thevenin’s resistance RTHfor given circuit is given by
Now consider the equivalent circuit (circuit (2)). The current through the load resistance I” in the
equivalent circuit is
Substituting for VTHand RTHfrom (5) and (6) in (7)
or
From equations (4) and (8), it is observed that I’ = I”.Hence Thevenin’s theorem is verified.
Maximum Power Transfer Theorem
Statement: The power transferred by a source to the load resistance in a network is maximum
when the load resistance is equal to the internal resistance of the source.
Proof of Maximum power transfer theorem:
Consider a network with a source of emfE and internal resistance r connected to a load resistance
RL. The current I in the circuit is
The power delivered to load resistance RL is
(from equation (1))
The variation of PLwith RLis as shown.
PLis found to be maximum for a particular value of RLwhen PLis maximum,

Differentiating
or
Thus
Thus the power delivered to the load resistance is maximum when the load resistance is equal to
the internal resistance of the source.
Show that the maximum power transfer efficiency of a circuit is 50%.
The power across the load
From the maximum power transfer theorem, PL is maximum when RL= r . Putting this condition
in equation (1),
The power that is taken from the voltage source is (or power generated by the source),
or
When RL = r

Dividing equation (2) by (3)
Thus the maximum power delivered to the load is only half the power generated by the source or
the maximum power transfer efficiency is 50%. The remaining 50% power is lost across the
internal resistance of the source.
Superposition theorem
Statement: In a linear network having number of voltage or current sources and resistances, the
current through any branch of the network is the algebraic sum of the currents due to each of the
sources when acting independently.
Explanation: By mesh current analysis
1. Consider the network as shown. The currents in different branches of the network are I1, I2and
I as shown. Also I1+ I2= I.
.
2. [Let the internal resistance r of the cells be negligible].
The cell E2is removed and the terminals are short as shown. Now the currents in the branches are
I1’, I2’ and I’. Also I1’ + I2’ = I’.
3. The E1is removed and the terminals are short as shown. The currents are I1’’, I2’’ and
I’’. Also I1’’ + I2’’ = I’’.
According to superposition theorem I1’+ I1’’ = I1, I2’+ I2’’ = I2and I’+ I’’ = I

Verification of superposition theorem:
1. Consider the network shown. Applying Kirchhoff’s voltage to the loop 1
2. Consider the circuit shown with E2 removed and terminals short. Applying Kirchhoff’s law to
loop 1
3. Consider the circuit with E1removed and terminals short. For loop (1)

Growth and Decay of Charge in C-R Circuit
Growth
Consider circuit containing resistor of resistance R and condenser of capacity C connected
in series with the battery of e.m.f. E through switch S, (position X) as shown in figure. At
any instant t
Voltage across resistor R = R i
Voltage across capacitor C = q/C
Now according to KVL,
q
E Ri
C
 
dq q
E R
dt C
 
dq
i
dt
 
  
q dq
E R
C dt
 

1dq
dt
q R
E
C

 
 
 
Integrating on both sides
1dq
dt
q R
E
C
 
 
 
 
 
1
ln
q
C E t K
C R
 
    
 
...... 1
Where K is constant of integration, it can be determine using initial condition, i.e. at t=0,
q=0.
lnC E K 
1n
eq 
1
ln ln
q
C E t C E
C R
 
    
 
1
ln ln
q
C E C E t
C R
 
    
 
1
ln
q
E
C t
E RC
  
  
   
  
   
1
q t
EC RC
  
Taking exponential on both sides
1
t
RC
q
e
EC

 
1
t
RC
q EC e
 
  
 
Let 0EC q be the maximum charge on the condenser.
0 1
t
RC
q q e
 
  
 
...... 2
Whent=T=RC (called time constant of the circuit)
1
0 0 0
1
1 1 0.63q q e q q
e
  
         
Thus time constant of RC circuit may be defined as the
time in which the charge on the capacitor grows to
63% of the maximum charge.
Now
0
0
t t
RC RC
qdq
i e i e
dt RC
 
  
This shows that as the charge on the capacitor
increases exponentially, the current in the circuit
decays exponentially from its initial value i0=E/R.

Decay
Consider circuit containing resistor of resistance R, condenser of capacity C as shown in
figure. Let the capacitor is charged to its maximum value (q0) when switch is at position X.
Suddenly at t = 0 switch is moved to connect at position Y, capacitor starts discharging. At
any instant t
0
q
Ri
C
 
0
dq q
R
dt C
 
dq
i
dt
 
  
q dq
R
C dt
 
1dq
dt
q RC
 
1dq
dt
q RC
   
1
ln q t K
RC
   ...... 1
Where K is constant of integration, it can be determine using initial condition, i.e. at t=0,
q=q0.
0ln q K 
1n
eq  0
1
ln lnq t q
RC
  
0
1
ln
q
t
q RC
 
0
t
RC
q q e

 ...... 2
This indicates exponential discharge of the condenser.
When t=T=RC (called time constant of the circuit)
1
0 00.37q q e q
 
Now
0
0
t t
RC RC
qdq
i e i e
dt RC
 
    
This shows that current decreases exponentially to zero.

Growth and Decay of Current in L-R Circuit
Growth
Consider circuit containing resistor of resistance R and inductor (coil) of inductance L
connected in series with the battery of e.m.f. E through switch S, (position X) as shown in
figure. Initially when switch is open there is no current in the circuit. Suddenly at t = 0
switch is moved to connect at position X, current starts to increase through inductor. It
gives rise to the opposing e.m.f. –L (di/dt) across inductor. At any instant t
Voltage across inductorL = L (di/dt)
di
E Ri L
dt
 
( )
di
E Ri L
dt
 
( )
dt di
L E Ri


1
( )
di
dt
L E Ri
 
 
 
1 1
ln E Ri t K
R L
    ...... 1
Where K is constant of integration, it can be determine using initial condition, i.e. at t=0, i=0.
1
ln E K
R
 
1n
eq   
1 1 1
ln lnE Ri t E
R L R
   
 
1 1 1
ln lnE Ri E t
R R L
   
1 1
ln
E Ri
t
R E L
 
  
 
ln 1
Ri R
t
E L
 
   
 
1
R
t
L
Ri
e
E

 
Let 0
E
i
R
 be the steady state value of current.
0
1
R
t
L
i
e
i

  ...... 2
0 1
R
t
L
i i e
 
  
 
When t=T=L/R (called time constant of the circuit)

1
0 0 0
1
1 1 0.63i i e i i
e
  
         
Thus time constant of LR circuit may be defined as the time in which the transient current
grows to 63% of its maximum value.
Figure shows plot of the current versus time. The current gradually rises from t=0
and attains maximum value i0after long time. And at t=T=L/R, the current is 0.63 i0.
Decay
Consider circuit containing resistor of resistance R,inductor (coil) of inductance Las shown
in figure. Let the current through the circuit reached to its steady state valuei0. Suddenly at
t = 0 switch is moved to connect at position Y and current starts to decrease from i0.At any
instant t
Voltage across inductorL = L (di/dt)
0
di
Ri L
dt
 
R di
dt
L i


R di
dt
L i

  
ln
R
i t K
L

  ...... 1
Where K is constant of integration, it can be determine using initial condition, i.e. at t=0, i=i0
0lni K 
1n
eq  0ln ln
R
i t i
L

 
0
0
ln ln
ln
R
i i t
L
i R
t
i L

 


0
R
t
L
i i e

 ...... 2
When t=T=L/R (called time constant of the circuit)
1
0 00.37i i e i
 
Figure shows plot of the current versus time. The current exponentially decay from t=0.
And at t=T=L/R, the current is 0.37i0.

Growth of Charge in Series LCR Circuit
Consider circuit containing resistor of resistance R,inductor (coil) of inductance L and
condenser of capacity C connected in series with the battery of e.m.f. E through switch S,
(position X) as shown in figure. Initially when switch is open there is no current in the
circuit. Suddenly at t = 0 switch is moved to connect at position X, current starts to increase
in the circuit. At any instant t
Voltage across inductor L = L (di/dt)
di q
E Ri L
dt C
  
2
2
dq di d q
i
dt dt dt
 
   
 
2
2
dq d q q
E R L
dt dt C
  
Divide by L on both side and rearrange
2
2
d q R dq q E
dt L dt CL L
    
Put, 21
2 and
R
b k
L LC
 
2
2 2
2
2
d q dq EC
b k q k EC
dt dt LC
      0As EC q
2
2 2
02
2
d q dq
b k q k q
dt dt
   
2
2
02
2 ( ) 0
d q dq
b k q q
dt dt
    
Put,
2 2
0 2 2
and
dx dq d x d q
x q q
dt dt dt dt
    
2
2
2
2 0
d x dx
b k x
dt dt
   
2 2
( 2 ) 0D bD k x   
This is second order differential equation. The auxiliary equation is
2 2
2 2
2 2
2 0 having two roots
2 4 4
2
m bm k
b b k
m b b k
  
  
    
Therefore solution of the equation may be written as
2 2 2 2
2 2 2 2
2 2
( ) ( )
( ) ( )
( )
b b k t b b k t
bt b k t bt b k t
bt pt pt p b k
x Ae Be
x Ae e Be e
x e Ae Be
     
    
   
 
 
 
0 ( ) ...........1bt pt pt
q q e Ae Be 
  
( ) ( ) ........2bt pt pt bt pt ptdq
i e Ape Bpe be Ae Be
dt
   
    

Where A and B are constants, they can be determine using initial condition, i.e. at t=0, q=0
in equation 1 and at t = 0, i = 0 in equation 2.
0 ( ) ...........3q A B  
And
0 ( ) ( )
0 ( ) ( ) ........4
Ap Bp b A B
p b A p b B
   
   
By solving equation 3 and 4 we get
{Multiply eqn 3 by (p+b) and add with eqn 4}
0
0
0 0
( ) ( ) ( )
( ) 2
( )
1
2 2
q p b p b A p b A
q p b pA
q p b q b
A
p p
     
  
 
      
 
Similarly we can obtain
0
1
2
q b
B
p
 
   
 
1n
eq 
0 0
0
0
0
0
1 1
2 2
1 1
2
1
1 1 1
2
bt pt pt
bt pt pt
bt pt pt
q qb b
q q e e e
p p
q b b
q q e e e
p p
b b
q q e e e
p p
 
 
 
    
         
    
    
        
    
      
         
      
Here three cases can be considered:
CASE 1:
2
2 2
2
2 2
1
i.e. >
4
is real
R
b k
L LC
p b k

 
It shows charging of capacitor in exponential manner, shown by curve A in figure. This is
called over damped.
CASE 2:
2
2 2
2
1
i.e. =
4
Roots are equal.
R
b k
L LC

It also shows the charging of capacitor shown by curve B in figure. This is called critically
damped.
CASE 3:
2
2 2
2
2 2
1
i.e. <
4
=j is imaginary
R
b k
L LC
p b k 

 

0
1
1 1 1
2
bt j t j tb b
q q e e e
j j
 
 
 
      
         
      
   0
1
1
2
bt j t j t j t j tb
q q e e e e e
j
   

    
      
  
0 1 cos sinbt b
q q e t t 

  
      
 0 1 cos sin
bt
e
q q t b t  


 
   
 
 0 1 cos cos sin sin
bt
e
q q k t k t   


 
   
 
Where sinθ=b/k and cosθ=ω/k
 0 1 cos( )btk
q q e t 

 
   
 
This equation represents damped oscillatory growth of charge, shown by curve C in figure.
The amplitude of cos term e-bt decreases exponentially with time.
Decay of Charge in Series LCR Circuit
Consider circuit containing resistor of resistance R, inductor (coil) of inductance L and
condenser of capacity C as shown in figure. Let the capacitor is charged to its maximum
value (q0) when switch is at position X. Suddenly at t = 0 switch is moved to connect at
position Y. At any instant t
Voltage across inductor L = L (di/dt)
0
di q
Ri L
dt C
  
2
2
dq di d q
i
dt dt dt
 
   
 
2
2
0
dq d q q
R L
dt dt C
  
Divide by L on both side and rearrange
2
2
0
d q R dq q
dt L dt CL
    
Put, 21
2 and
R
b k
L LC
 
2
2
2
2 0
d q dq
b k q
dt dt
   
Put,
2 2
( 2 ) 0D bD k q   
This is second order differential equation. The auxiliary equation is

2 2
2 2
2 2
2 0 having two roots
2 4 4
2
m bm k
b b k
m b b k
  
  
    
Therefore solution of the equation may be written as
2 2 2 2
2 2 2 2
2 2
( ) ( )
( ) ( )
( )
b b k t b b k t
bt b k t bt b k t
bt pt pt p b k
x Ae Be
x Ae e Be e
x e Ae Be
     
    
   
 
 
 
( ) ...........1bt pt pt
q e Ae Be 
 
( ) ( ) ........2bt pt pt bt pt ptdq
i e Ape Bpe be Ae Be
dt
   
    
Where A and B are constants, they can be determine using initial condition, i.e. at t=0, q=q0
in equation 1 and at t=0, i=0 in equation 2.
0 ( ) ...........3q A B  
And
0 ( ) ( )
0 ( ) ( ) ........4
Ap Bp b A B
p b A p b B
   
   
By solving equation 3 and 4 we get
{Multiply eqn 3 by (p+b) and add with eqn 4}
0
0
0 0
( ) ( ) ( )
( ) 2
( )
1
2 2
q p b p b A p b A
q p b pA
q p b q b
A
p p
    
 
 
    
 
Similarly we can obtain
0
1
2
q b
B
p
 
  
 
1n
eq 
0 0
0
1 1
2 2
1 1
2
bt pt pt
bt pt pt
q qb b
q e e e
p p
q b b
q e e e
p p
 
 
    
       
    
    
       
    
Here three cases can be considered:
CASE 1:
2 2
2
2 2
2
1
i.e. >
4
is realp b k
R
b k
L LC
 

This shows discharge of capacitor to zero. This is called over damped. This is shown by
curve A in figure.

CASE 2:
2
2 2
2
1
i.e. =
4
Roots are equal.
R
b k
L LC

This is called critically damped. This is shown by curve B in figure.
CASE 3:
2
2 2
2
2 2
1
i.e. <
4
=j is imaginary
R
b k
L LC
p b k 

 
0
1 1
2
bt j t j tq b b
q e e e
j j
 
 
     
       
    
   0
2
bt j t j t j t j tq b
q e e e e e
j
   

   
    
 
0 cos sinbt b
q q e t t 

  
   
 0 cos sin
bt
e
q q t b t  


 
 0 cos cos sin sin
bt
e
q q k t k t   


 
Where sinθ=b/k and cosθ=ω/k
 0 cos( )btk
q q e t 


 
This shows oscillatory discharge.
The amplitude of sin term e-bt decreases exponentially with time.
sinbtE
i e t
L



 
Ballistic galvanometer
A galvanometer is an instrument, which uses magnetic effects for detecting and
measuring currents or electric charge. In a moving coil galvanometer, a flat conducting coil
is suspended between the poles of a permanent magnet (fig a). The coil consists of an
insulated wire wound on light brass or aluminum frame.
The coil is usually suspended by a phosphor bronze strip, which also serves as the
current lead for the current to the coil and is finally connected to a terminal at the base of
the instrument. The other end of the coil is connected to a light spring, which provides the
restoring couple. The other end of the spring is attached to the other terminal.

fig: a
A ballistic galvanometer is used for the purpose of measuring the total charge in an
impulse current instead of measurement of a steady current. For this we require that the
period of oscillation of the moving coil be large compared to the time for the current pulse
passing through it. This is achieved by loading the coil so as to increase its moment of
inertia. Thus when the current pulse passes through it, the coil receives a kick due to an
impulse torque. Subsequently the coil oscillates freely due to the restoring torque provided
by the suspension. The maximum deflection amplitude 0 (called first throw position) is
reached long after the current is passed. A necessary requirement is also that damping is
small.
fig:b
The deflection of the coil is measured by the reflection of a light beam by the mirror
attached to the coil (fig b). This deflection is recorded as the linear deflection of the spot, X,
that the reflected beam makes on the scale. Clearly X will also be directly proportional to
the charge q discharged through the ballistic galvanometer.
Theory
The first throw position 0 (maximum amplitude of deflection) is proportional to
the total charge q i dt  in the current i passing through the coil.
If A is the area of the coil, n is the number of turns of the coil and B is the magnetic
field of the permanent magnet. Then the torque acting on the coil is
niAB 
It will give rise to initial angular momentum
I dt nAB idt nABq    
(I is the moment of inertia of the coil)
I
q
nAB



If the restoring torque provided by suspension is 0C then equating the initial
kinetic energy to the work done by the restoring torque, we get
2 2
0
2 2 2
0
0
1 1
2 2
I C
I IC
I IC
 
 
 

 
 
which leads to (using eq 5 ),
0
IC
q
nAB

Substituting, 2
I
T
C
 the period of free oscillations, we get
0
I
Cq C
nAB

0
2
T C
q
nAB



Showing the proportionality relationship between q and 0.
0 0OR '
2
T
q K q K 

 
is called current reduction factor i.e. current required to produce unit deflection.
' is called ballastic reduction factor i.e. charge required to produce unit deflection.
2
C
K
nAB
T C
K
nAB


Charge Sensitivity of Ballistic Galvanometer:
Charge Sensitivity (Sq) of Ballistic Galvanometer is defined as the throw in
millimeters on a scale one meter apart when charge of one micro coulomb passes through
it.
0 2 2
arg
Deflection nAB
Sq Si
Ch e q T C T
  
   
Si is called current sensitivity of galvanometer.

B.Sc. Sem II Network theorems

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