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Chapter 1
Direct-Current Circuits

DEFINATIONS
Linear elements : In an electric circuit, a linear element is an electrical
element with a linear relationship between current and voltage
...

Nonlinear Elements :A nonlinear element is one which does not have a
linear input/output relation
...
Most semiconductor devices have non-linear
characteristics
...
Some of the examples are batteries,
generators , transistors, operational amplifiers , vacuum tubes etc
...


In unilateral element, voltage – current relation is not same for both
the direction
...

In bilateral element, voltage – current relation is same for both the
direction
...
It is only a function of time
...

Ideal Current Source: The current generated by the source does not
vary with any circuit quantity
...
Such a source
is called as an ideal current source
...
The resistance of element is denoted by the symbol
“R”
...


• Electric Current
• Resistance and Ohm’s Law
• Energy and Power in Electric Circuits
• Resistors in Series and Parallel
• Kirchhoff’s Rules
• Circuits Containing Capacitors
• RC Circuits
• Ammeters and Voltmeters

Electric Current
Electric current is the flow of electric charge from
one place to another
...


Electric Current
A battery uses chemical reactions to produce a
potential difference between its terminals
...


Electric Current
A battery that is disconnected from any circuit
has an electric potential difference between its
terminals that is called the electromotive force or
emf:
Remember – despite its name, the emf is an
electric potential, not a force
...
Therefore, current flows
around a circuit in the direction a positive charge
would move;
electrons move
the other way
...


Electric Current
Finally, the actual motion of electrons along a
wire is quite slow; the electrons spend most of
their time bouncing around randomly, and have
only a small velocity component opposite to
the direction of the current
...

Ohm’s law relates the voltage to the current:

Be careful – Ohm’s law is not a universal law
and is only useful for certain materials
(which include most metallic conductors)
...
This property of a material is
called the resistivity
...
This property can be used in
thermometers
...


3 Energy and Power in Electric Circuits
When a charge moves across a potential
difference, its potential energy changes:

Therefore, the power it takes to do this is

3 Energy and Power in Electric Circuits
In materials for which Ohm’s law holds, the
power can also be written:

This power mostly becomes heat inside the
resistive material
...

They are charging you for energy use, and kWh
are a measure of energy
...
They can be replaced by a single
equivalent resistance without changing the
current in the circuit
...
Replace these with their
equivalent resistances; as you go on you will be
able to replace more and more of them
...

For these circuits, Kirchhoff’s rules are useful
...


5 Kirchhoff’s Rules
The junction rule: At any junction, the current
entering the junction must equal the current
leaving it
...


5 Kirchhoff’s Rules
Using Kirchhoff’s rules:
• The variables for which you are solving are the
currents through the resistors
...

• You will need both loop and junction rules
...

When capacitors are
connected in parallel,
the potential difference
across each one is the
same
...

The total potential
difference is the sum of the
potential differences
across each one
...


7 RC Circuits
In a circuit containing
only batteries and
capacitors, charge
appears almost
instantaneously on the
capacitors when the
circuit is connected
...


7 RC Circuits
Using calculus, it can be shown that the charge
on the capacitor increases as:

Here, τ is the time constant of the circuit:

And

is the final charge on the capacitor, Q
...
time for an RC circuit:

7 RC Circuits
It can be shown that the current in the circuit
has a related behavior:

8 Ammeters and Voltmeters
An ammeter is a device for measuring current,
and a voltmeter measures voltages
...


8 Ammeters and Voltmeters
A voltmeter measures the potential
drop between two points in a circuit
...


Transients Analysis

Solution to First Order Differential Equation
Consider the general Equation
dx(t )

 x(t )  K s f (t )
dt

Let the initial condition be x(t = 0) = x( 0 ),
then we solve the differential equation:

dx(t )

 x(t )  K s f (t )
dt
The complete solution consists of two parts:
• the homogeneous solution (natural solution)
• the particular solution (forced solution)

The Natural Response
Consider the general Equation
dx(t )

 x(t )  K s f (t )
dt

Setting the excitation f (t) equal to zero,
dxN (t )
dxN (t )
x N (t ) dxN (t )
dt

 x N (t )  0 or

,

dt
dt

x N (t )

dxN (t )
dt
  ,

x N (t )


x N (t )   e t / 

It is called the natural response
...


The Complete Response
Solve
Consider the general Equation for ,
dx(t )

 x(t )  K s f (t )
dt

The complete
response is:
• the natural response +
• the forced response
x  x N (t )  xF (t )

  e t /   K S F
  e  t /   x ( )

for t  0
x(t  0)  x(0)    x()

  x(0)  x()

The Complete
solution:

x(t )  [ x(0)  x()]et /   x()

[ x(0)  x()]et /  called transient respon
x() called steady state response

WHAT IS TRANSIENT RESPONSE

Figure 5
...
Solve first-order RC or RL circuits
...
Understand the concepts of transient
response and steady-state response
...
Relate the transient response of firstorder
circuits to the time constant
...
It
is called natural response
...


(source free response)

Discharge of a Capacitance through a
Resistance

ic

iR

i  0,

iC  iR  0

dvC t  vC t 
C

0
dt
R

Solving the above equa
with the initial condition
Vc(0) = Vi

Discharge of a Capacitance through a
Resistance
1

dvC t  vC t 
C

0
dt
R

dvC t 
RC
 vC t   0
dt
vC t   Ke

RC

vC t   Ke

t RC



vC (0 )  Vi
 Ke

st

RCKse  Ke  0
st

s

st

0 / RC

K

vC t   Vi e

 t RC

vC t   Vi e t

RC

Exponential decay waveform
RC is called the time constant
...
8%
of the initial voltage
...

At time constant, the
voltage is 63
...


Figure
5
...
18

a)
...
Complete, natural, and
forced responses of the
circuit
5-8

Circuit Analysis for RC Circuit
iR

iC

R
Vs

Apply KCL

+ VR -

C

+
Vc
-

iR  iC
vs  v R
dvC
iR 
, iC  C
R
dt
dvC
1
1

vR 
vs
dt
RC
RC

vs is the source applied
...


The Forced Response
Consider the general Equation
dx(t )

 x(t )  K s f (t )
dt

Setting the excitation f (t) equal to F, a
constant for t 0

dxF (t )

 xF (t )  K S F
dt
x F (t )  K S F for t  0

It is called the forced response
...
01
microF

iC
+
Vc
-

Initial condition Vc(0) = 0V
iR  iC
vs  vC
dvC
iR 
, iC  C
R
dt
dvC
RC
 vC  vs
dt
5
 6 dvC
10  0
...
01
microF

+
Vc
-

dx(t )

 x(t )  K s f (t )
dt
x  x N (t )  xF (t )
 e
 e

t / 



vc  100  Ae

t
103

As vc (0)  0, 0  100  A

and
t / 

Initial condition Vc(0) = 0V
3 dvC
10
 vC  100
dt

 KS F
 x ( )

A  100
vc  100  100e



t
103

Energy stored in capacitor
dv
p  vi  Cv
dt
t
t o

pdt



dv
t
 t Cv dt
o
dt

 C tt vdv

1
2
2
 C v(t )  v(to )
2



o

If the zero-energy reference is selected at to, im
capacitor voltage is also zero at that instant, th

1 2
wc (t )  Cv
2

RC CIRCUIT
R

C

Power dissipation in the
resistor is:

pR = V2/R = (Vo2 /R) e -2 t /RC

Total energy turned into heat in
2  2t / RC
the resistor
V
e
dt
o 0

WR  0 p R dt 
2
 Vo R(

1
2
 CVo
2

R

1
 2t / RC 
)e
|0
2 RC

RL CIRCUITS
Initial condition
i(t = 0) = Io

i(t)

VR
+

R

L

di
vR  vL  0  Ri  L
dt
+
L di
VL
i  0
R dt
Solving the differenti equation
al

RL CIRCUITS

VR
+

di R
 i0
dt L
i(t)
i ( t ) di
di
R
  dt,

Io
i
L
i
+
R t
R
L
VL
i
ln i |I o   t |o
L
R
ln i  ln I o   t
Initial condition
L
 Rt / L
i(t = 0) = Io
i (t )  I o e





t

R
 dt
o
L

RL CIRCUIT
Power dissipation in the
resistor is:
i(t)

VR
+

2e-2Rt/L
pR = i2R = Ioturned R heat in
Total energy
into

the resistor
2
WR  p R dt  I o R
+

R

L

VL
-



0



2
I o R( 





e 2 Rt / L dt

0

L 2 Rt / L 
)e
|0
2R

1 2
 LI o
2
1 2
LI o
It is expected as the energy stored in the inductor
2

i(t)

Vu(t)
+
_

Vu(t)

R

L

+
VL
-

RL CIRCUIT

di
Ri  L  V
dt
Ldi
 dt
V  Ri
Integrating both sides,
L
 ln(V  Ri )  t  k
R

L
i (0 )  0, thus k   ln V
R
L
 [ln(V  Ri )  ln V ]  t
R
V  Ri
 e  Rt / L
or
V
V V  Rt / L
i  e
, for t  0
R R


where L/R is the time consta



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