ac circuits - ek
TRANSCRIPT
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAOLU, Page 1
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AC Circuits
by
Prof. Dr. Osman SEVAOLU
Electrical and Electronics Engineering Department
Alternating Current AC) Circuits
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AC Circuits
What is Direct Current DC) ?
Direct Current (DC) is a current with aconstant time characteristics
Definition
Current, I
+
R1= 5 Ohms
Vs=600 V
Switch
R2= 5 Ohms
Switch is turned on at: t = 1 sec
1 2 3 4 5 6
80
0
Current(A
mp)
7
20
40
60
I = 60 A
DC (Constant) Current
Time (Sec)
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AC Circuits
What is Alternating Current AC) ?
Alternating Current (AC) is a current with
time varying characteristics
Definition
Sinusoidal AC
Non - Sinusoidal AC
0,0
1,0
2,0
3,0
4,0
5,0
0,0 1,0 2,0 3,0 4,0 5,0 6,0t (sec)
Current(Am
p)
10
- 10
0
Angle (Radians)Current(Am
p)
/2 3
/2 2
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AC Circuits
Basic Parameters of a Sinusoidal Waveform
Sinusoidal voltage is a voltagewith waveform as shown on theRHS
Definition
Sinusoidal Voltage
V(t) = V sin ( wt + )^
where
V(t) is the voltage waveform,
V is the peak value (amplitude),
w is the angular frequency, is the phase shift, i.e. angle of the
voltage at t = 0, (phase angle)
^
w = 2 f
f = 50 Hz
100
0
Angle (Radians)
Voltage (Volt)
/2 3/2 2
200
300
312
Amplitude
Phase angle
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AC Circuits
AC Alternating Current) Circuit
Positive half cycle
Negative half cycle
-25
-20
-15
-10
-5
0
5
10
15
20
25
0 0,005
Load
+
Load0,015
I (Amp)
-25
-20
-15
-10
-5
0
5
10
15
20
25
0 0,005 0,01
I (Amp)
-25
-20
-15
-10
-5
0
5
10
15
20
25
0,01 0,015 0,02
+V(t)
I(t)
V(t)
I(t)
Time (Sec) Time (Sec)
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AC Circuits
Capacitance
Capacitors store electrical charge
Definition
Storage capacity of a capacitor is
called capacitance +
_
+
_
Small Capacitance Large Capacitance
Water (hydroulic) example
Capacitance = C1 Capacitance = C2C1 < C2
_
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AC Circuits
Geometry
Control relay
Capacitor banks
Capacitor-Practical Configuration
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AC Circuits
MV Medium Voltage) Shunt Capacitor Banks
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AC Circuits
MV Medium Voltage) Shunt Capacitor Banks
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AC Circuits
Electronic Capacitors in a Motherboard
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AC Circuits
Basic Relation
Charge stored in a capacitor is
proportional to the capacitance C,Charge stored in a capacitor is
proportional to the voltage Vapplied
Basic Principle
or
Q = C Vwhere, Q is charge stored (Coulombs),
V is voltage (Volts),C is capacitance (Farads) +
_
+
1 Volt
Symbolic Representation
C = 1 Farad
+
_
+
VC
Q = 1 Coulomb1 Farad is the capacitance with a
charge of 1 Coulomb at a voltage 1 Volt
between the plates
Capacitance CVoltage Source V
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AC Circuits
Voltage - Current Relation for a Capacitor
Definition
Q = C V
The relation;
may be written in time domain as;
Q(t) = C V(t)
or differentiating both sides with respect to time
dQ(t)/dt = C dV(t)/dt
or remembering thatdQ(t)/dt = I(t)
Hence,
I(t) = C dV(t) / dt
+
_
C
I(t)
V(t)
+
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AC Circuits
Definition
The above equation may be integratedwith respect to time, yielding the
following Voltage - Current Relation for a
Capacitor
V(t) = (1/C) I(t)dt + V(0)
where V(0) is the initial voltage across
the capacitor, representing the initialvoltage due to the initial charge stored in
the capacitor
I(t)
+
_
+ C
V(t)
Switch
Switch is turned on at: t = 0 sec
Vc (0)
Voltage - Current Relation for a Capacitor
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AC Circuits
Example - 1
Problem
Calculate the time waveform of the current flowingin the circuit shown on the RHS by assuming that
the capacitor is charged by the exponential
voltage V(t) shown in the figure
V(t) = 5 (1 - e-t/)
where is known as the time constant of the
circuit and given as = 10 -6sec = 1.0 sec
C = 0.1 F
+
_
+
t (sec)
V(t) = 5 (1 - e-t/
)
0,0
1,0
2,0
3,0
4,0
5,0
0,0 1,0 2,0 3,0 4,0 5,0 6,0
C
I(t)
V(t)
Vmax = Maximum voltage that can be reached = 5 Volts
Qmax = C x Vmax = Maximum charge that can be stored
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AC Circuits
Example - 1
Solution
0,0
1,0
2,0
3,0
4,0
5,0
0,0 1,0 2,0 3,0 4,0 5,0 6,0
t (sec)
I(t) = C dV(t)/dt
V(t) = 5 (1 - e-t/)
Hence,
I(t) = C d V(t) / dt
= C d 5(1 e -t/)/dt
= 0.1 x10-6( 5 /
) xe -t/
= 0.1 x10-6x5 x106x e -t/
= 0.5 xe -t/ Ampers
0,00,0 1,0 2,0 3,0 4,0 5,0 6,0
t (sec)
0,5
0,1
0,2
0,3
0,4
I(t)
+
_
+ C
I(t)
V(t)
V(t) = 5 (1 - e-t/ )
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AC Circuits
Example - 2
Problem
Current source shown in the circuit onthe RHS provides a 10 mA constant
current within the time interval;
Capacitor is initially charged to 2 Volts
voltage
t [0, 1]
Determine the voltage across thecapacitor within the time interval;
t [0, 1]
I(t) (mA)
10,0
0,0 0,2 0,4 0,6 0,8 1,0 1,2
t (sec)
8,0
6,0
4,0
2,0
12,0
14,0
C = 1000 F Vc(0) = 2 Volts
+
_
I(t)
I(t)C
Switch
Switch is turned on at: t = 0 sec,off at: t = 1.0 sec.
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AC Circuits
Example - 2
Solution
Voltage across the capacitance isexpressed as
t [0, 1]
V(t) = (1/C) I(t)dt + V(0)where,
V(0) = 2 Voltsis the initial voltage across thecapacitorHence;
C = 1000 F
Vc(0) = 2 Volts
I(t) (mA)
+
_
I(t)
I(t)C
0,0
10,0
0,0 0,2 0,4 0,6 0,8 1,0 1,2
t (sec)
8,0
6,0
4,0
2,0
12,0
14,0
V(t) (Volts)
0,0
10,0
0,0 0,2 0,4 0,6 0,8 1,0 1,2
t (sec)
8,0
6,0
4,0
2,0
12,0
14,0
1
V(t) = (1/C)
I(t)dt + 20
1
V(t) = (1/10-3) 10-2 dt + 20
= 10 t + 2 Volts
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AC Circuits
R-C Circuits
Problem
Solve the RC circuit shown on the RHS
for current I(t) flowing in the circuit whenthe switch is closed at t = 0
Differentiating both sides wrt time once;
dV(t) / dt = R dI(t) / dt + (1/C) I(t)or
d/dt I(t) + (1/RC) I(t) = (1/R) dV(t) / dt
A first order ordinary differential equation
V(t) = R I(t) + VC(t)
= R I(t) + (1/C) I(t)dt + V(0)
Writing down KVL for the circuit
Solution R
C
+
_
+
I(t)
V(0)=V0
V(t) = V sinwt^
Switch is turned on at: t = 0 sec
Switch
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AC Circuits
Solution of the resulting First Order Ordinary Differential Equation
Solve the resulting first order ordinary differential equation (ODE)
Solution
dI(t) / dt + (1/RC) I(t) = (1/R) dV(t) / dt
dI(t) / dt + (1/RC) I(t) = (1/R) d/dt (V sin ( wt + ))
dI(t) / dt + (1/RC) I(t) = (V/R) w cos ( wt + ) V( t ) = V sin ( wt + )dV( t ) / dt = V w cos wt
^^
100
0
Voltage (Volt)
Angle (Radians)
/2 3/2 2
200
300
312
Amplitude
Phase angle
^
^
R
C
+
_
+ V(0)=V0
V(t) = V sinwt^
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AC Circuits
Solve the resulting first order ordinary differential equation (ODE)
Solution
(t) dI(t)/dt+ I(t) (1/RC)(t) = (t) ( V/R ) w cos ( wt + )
(t) dI(t)/dt+ I(t) d/dt (t) = ( V/R ) (t) w cos ( wt + )
d/dt [(t) I(t)] = ( V/R ) (t) w cos ( wt + )
d/dt [(t) I(t)] dt = ( V/R ) (t) w cos ( wt + ) dt + I(0)
(t) I(t) = ( V/R ) (t) w cos ( wt + ) dt + I(0)
I(t) = I (t) -1
(t) w cos(wt + ) dt + (t)-1I(0)
^
^
^
^
^
^
Define an integration factor(t) = e t / R C
Multiply both sides of the above ODE by this factor;
V( t ) = V sin ( wt + )dV( t ) / dt = V w cos wt
^
^
100
0
Voltage (Volt)
Angle (Radians)
/2 3/2 2
200
300
312
Amplitude
Phase angle
dI(t) / dt + (1/RC) I(t) = (V/R) w cos ( wt + )^
Solution of the resulting First Order Ordinary Differential Equation
R
C
+
_
+ V(0)=V0
V(t) = V sinwt^
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AC Circuits
Solution of the resulting First Order Ordinary Differential Equation
Solution Continued)
Subsituting the above term into the solution;I(t)
V( t ) = V sin wt
dV( t ) / dt = V w cos wt
^
^
Switch is turned on at: t = 0 sec
R
C
+
_
+ V(0)=V0
V(t) = V sinwt^
w sinwt + (1/RC) coswtI(t) = I e - t/RCwe t/RC ---------------------------------- + e - t/RCI(0)
(1/RC)2+ w 2
^
sinwt + (1/wRC) coswtI(t) = I w2----------------------------------- + e - t/RCI(0)
(1/RC)2+ w 2
^
II(t) = -------------------(sinwt + (1/wRC) coswt) + e - t/RCI(0)
(1/wRC)2+ 1
^
Steady-State Term Transient Term
Switch
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AC Circuits
Numerical Example
Now assume that the parameters of the circuit on
the RHS are as follows;
V(t) = 312 sin wt Volts
R = 10 Ohms
C = 10 Farads
I(t)
R
C
+
_
+ V(0)=V0
V(t) = V sinwt^
V( t ) = V sin wt
dV( t ) / dt = V w cos wt
^
^
Switch
Switch is turned on at: t = 0 sec
I
I(t) = -------------------(sinwt + (1/wRC) coswt) + e - t/RCI(0)(1/wRC)2+ 1
^
Steady-State Term Transient Term
Solution of the resulting First Order Ordinary Differential Equation
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AC Circuits
Numerical Example
I
I(t) = -------------------(sinwt + (1/wRC) coswt) + e - t/RCI(0)(1/wRC)2+ 1
^
Steady-State Term Transient Term
-20-20
-15
-10
-5
0
5
10
15
20
25
0.5 1 1.5 2 2.5 3 3.5
5
10
15
20
25
0 0.5 1 1.5 2 2.5 3 3.5 4
-30
-20
-10
0
10
20
30
40
0.5 1 1.5 2 2.5 3 3.5 4
Solution of the resulting First Order Ordinary Differential Equation
AC Circuits
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Solution for DC Voltage Source - Two Simple Rules
Rule - 1
A capacitor acts as a DC voltage source
initially due to its initial charge
I(0) = ( V V0) / R
The current waveform will then be;
I(t) = I() + [ I(0) - I() ] e -t/
= RC = Time Constant: The time required a for
capacitor to reach 63 % of its full charge
= 2 x 1000 F = 2 x 1000 x 10-6 = 0.002 sec
Then the initial value of current will be;
I(0) = ( VDC V0) / R
C =1000 F+
_
+
I(t)VDC
R = 2
V(0)=V0V = VDC-V0
R
+
_
+
I(0) SC
0
V(t) = (1 / C) I(t)dt = V(0) = V0-
Please note that anuncharged capacitoracts effectively as a
short circuit,i.e. V0= 0
AC Circuits
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Solution for DC Voltage Source - Two Simple Rules
Rule - 2
A capacitor acts as an open circuit finally
to a DC voltage (or current)
The current waveform will then be;
I(t) = I() + [ I(0) - I() ] e -t/
I() = 0
I(t) = C d/dt V(t) = C d/dt (constant) = 0 (OC)
+
_
+
I(t)VDC
V(0)=V0
OC
+
I() = 0
R
VDC
Then the final value of current will be;
I() = 0
C =1000 F
R = 2
= RC = Time Constant: The time required for
a capacitor to reach 63 % of its full charge
= 2 x 1000 F = 2 x 1000 x 10-6 = 0.002 sec
AC Circuits
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Solution for DC Voltage Source - Two Simple Rules
Solution
Substituting the above expressions into
the current expression
The current waveform will then be;
I(t) = I() + [ I(0) - I() ] e -t/
I(t) = [( VDC V0) / R ] e-t/
The initial value of current will be;
I(0) = ( VDC V0) / R
C =1000 F+
_
+
I(t)VDC
R = 2
V(0)=V0
The final value of current will be;
I() = 0
I(t) (Amp)
0
4.0
8.0
12.0
16.0
20.0
24.0
0 0,002 0,004 0,006 0,008 0,01
% 63 of the initial value
AC Circuits
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Meaning of Time Constant
Definition
Time constant of a electric circuit is
the duration for the current to getreduced by 63 % of its initial value
Time constant
of an RC circuit issimply expressed as:
= RC
The Effect of
0
0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 5 5,5 6
t (sec)
1
2
3
4
5
% 63 of the initial value
% 63 of the inital value
1
2
2 >
1
AC Circuits
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Example
Problem
Find the voltage waveform across the 1 mF
capacitor shown on the RHS, when it has aninitial charge of 6 Volts and charged by a 24Volts DC voltage source through a wire with 2Ohm resistance
C = 1 mF
R= 2 Ohm
+
_
+
I(t)
VDC= 24 Volts V(0)=V0 = 6 V
The voltage waveform will then be;
V(t) = V() + [V(0) - V()] e -t/
V(t) = 24 + ( 6 24 ) e-t/0.002
= 24 -18 e-t/0.002
Volts
Capacitor will behave as a DC source at thebeginning and as OC at the end, hence;
V(0) = V0= 6 Volts
V() = VS= 24 Voltst (sec)
V(t)(Volts)
0
4.0
8.0
12.0
16.0
20.0
24.0
0 0,002 0,004 0,006 0,008 0,01
% 63 of the final value
= RC = Time Constant: The time required
for a capacitor to reach 63 % of full charge
= 2 x 1000 F = 2 x 0.001 = 0.002 sec
AC Circuits
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Energy Stored in a Capacitor
Problem
Calculate the instantaneous energy
stored in a capacitor with a capacitanceC and an instantaneous voltage VC(t)
P(t) = VC(t) I(t)
WC(t) = P(t) dt
=
VC(t) I(t) dt
= VC(t) C dVC(t) / dt dt
= C
VC(t) dVC(t)
WC(t) = (1/2) C VC2
(t)
or
Example
A 10 F capacitor fully charged with a 12
Volts DC voltage stores an energy;(1/2) 10 x10-6x122 = 720 x10-6Joule
C+
_
+
I(t)
V(t) Vc
(t)
AC Circuits
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Example
Problem
Find the instantaneous energy in thecapacitor for the voltage shown in the
figure
Wc(t) = (1/2) 10 x 10-6 V 2sin 2wt
= 0,000005 x 3122sin2wt
= 0,4867 sin2wt= 0,4867 ( cos 2x )
V (t) = V sin wt
WC(t) = (1/2) C VC2(t)
^
^
V( t ) = V sin wt^
312
- 312
Angle (Radians)
0/2 3/2 2
Wc(t)
0,1
0,2
0,3
0,4
0,5
0,6
0/2 3/2 2
Voltage (Volts)
Angle (Radians)
C +
_
+
I(t)
C = 10 F
V(t) = 312 sin wt
Mean of Wc(t) > 0sin2wt = 1 cos2wt = 1 ( 1 + cos2wt ) / 2
= cos2wt
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AC Circuits
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Series Connected Capacitances
V1(t) = (1/C
1) I(t)dt
V2(t) = (1/C2) I(t)dt
----------- = ----------------------
V (t) = [ (1/C1) + (1/C2) ]
I(t)dt= (1/Ctot) I(t)dt
Hence,
Series connected capacitances
+
I(t)
C1
C2
V1 (t)
V2 (t)
+
+
+ +
1Ctot = ----------------------
(1/C1) + (1/C2)
Series connected capacitancesare combined in the same way asfor shunt connected resistances
V(t)
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AC Circuits
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Series and Shunt Connected Capacitances
Where,C tot= C1 + C2
is the total capacitance
Shunt connected capacitances aresimply added
Shunt connected capacitances
+
I(t)
V(t)C2C1
I2I1
Vc(t)
I1(t) = C1 d V(t) / dtI2(t) = C2d V(t) / dt
----------- = ----------------------------
I (t) = (C1 + C2) d V(t) / dt= Ctot d V(t) / dt
+ +
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AC Circuits
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Inductance
Inductance is a winding or
coil of wire around a core
Definition
Coil Core Toroidal Coil Toroidal Core
Core may be either insulator
or a ferromagnetic material
+
_
Symbolic representation
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AC Circuits
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Ferrite Core Toroidal Inductor
Ferriet core inductor has a
toroidal ferrit core inside
Definition
II
Toroidal coil
Ferrite core
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AC Circuits
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Air Core Inductor
Configuration Air core inductor has no core inside
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AC Circuits
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Basic Relation
Voltage across an inductor isproportional to the rate of change ofcurrent
Definition
V(t) = L d I(t) / dt
where, V(t) is the voltage across theinductance,I(t) is the current flowingthrough,L is the inductance (Henry)
1 Henry is a value of inductance defined as
1 Henry = 1 Voltx1 second / 1 Amp
Voltage Source V(t)
I(t)
Inductance L
+
+
_
V(t)L
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Current in an Inductance
The equation;
Definition
V(t) = L d I(t) / dt
can be written in inverse form as
I(t) = (1/L)
V(t)dt + I(0)
where I(0) is the current initially flowingin the inductor
Voltage Source V(t)
++
Inductance L
_
I(t)
V(t) L
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I (Amp)
V(t)
I(t)
Inductance+
Phase Shift between Current and Voltage Waveforms
I(t) = (1/L) V(t) dt
= (1/L)
Vmaxsin wt dt= - (Vmax/ wL) coswt
= - Imaxcoswt
+
-25
-20
-15-10
-5
0
5
10
15
20
Time (Sec)
25
0.005 0.010 0.015 0.020
V (Volts), I (Amp)
VmaxImax
Current in an Inductance
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Series and Shunt Connected Inductors
Series connected inductors are added
V1
(t) = L1
d I(t) / dt
V2(t) = L2d I(t) / dt
----------- --------------------------
V(t) = (L1
+ L2
) d I(t) / dt
= L tot d I(t) / dt
where
Ltot
= L1
+ L2
is the total inductance
+ + +
I(t)
V(t)
L2
L1
V2 (t)
V1 (t)+
+
Series connected inductances
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Series and Shunt Connected Inductors
I1(t) = (1/L1) V(t)dt
I2(t) = (1/L2) V(t)dt
-------- --------------------I(t) = [ (1/L1) + (1/L2) ] V(t)dt
= (1/Ltot) V(t)dt
+
Shunt connected inductances
+
I(t)
V(t)L
2
L1
I2I1
Vc(t)
1Ltot = ----------------------
(1/L1) + (1/L2)
Hence,
+
Shunt connected inductances arecombined in the same way as for shuntconnected resistances
+
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Example - 3
Problem
Calculate the voltage across the 10 mHinductor with the current shown in the figure
on the RHS
0,0 2,0 4,0 6,0
t (msec)8,0 10,0 12,0 14,0
0,0
0,5
1,0
I(t) = 0 t < 1 ms
I(t) = 1/((5-1)x10-3) (t 10-3) 1 t 5 msI(t) = 1 5 t 9 msI(t) = -1/((5-1)x10-3) (t 12 x10-3) 9 t 13 msV(t) = 0 t 13 ms
I(t) (Amp)
L = 10 mH
+
I(t)
I(t)
+
L
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Example - 3
Solution
V(t) = L d I(t) / dt
V(t) (Volts)
0,0
0,0 2,0 4,0 6,0
t (msec)
8,0 10,0 12,0 14,0
-2,5
2,5
V(t) = 0 t < 1 msV(t) = 0.01x1/((5-1)x10-3) = 2.5 V 1 t 5 msV(t) = 0 5 t 9 msV(t) = 0.01x(-1)/((5-1) x10-3) = -2.5 V 9 t 13 ms
V(t) = 0 t 13 ms
Differentiating the current waveform and
multiplying by L (L = 10-2H);
L = 10 mH
+
I(t)
I(t)
+
L
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Example - 4
Problem
Assume that the inductor shown on the RHS
is connected to a voltage source of the form
shown in the figure
Find out the inductor current waveform
assuming that the initial current in the
inductor is zero
V(t) = 0 t < 0 s
V(t) = -10 V 0 t 1 s
V(t) = 0 t 1 s
t (sec)
0,0
0,0
-10
0,5 1,0
V(t)
+
I(t)
+
L L = 10 mH
V(t) (mV)
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Example - 4
Solution
I(t) = (1/L) V(t) dt + I(0)
I(t) = 0 t < 0 s
I(t) = 1/(10x10-3)x V(t) dt= 100 x V(t) dt
= 100 x (-10 x 10-3
) t = - t 0 t 1 sI(t) = -1 A t 1 s
t (sec)
I(t) (A)
0,0
0,0
-1
0,5 1,0
V(t)
+
I(t)
+
L L = 10 mH
I(t) = (1/L) V(t) dt + I(0)
= 1/(10 x 10-3)
V(t) dt
Integrating the voltage expression;
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Energy Stored in an Inductor
Problem
Calculate the instantaneous energy
stored in an inductor with an inductanceL and an instantaneous voltage VL(t)
P(t) = VL(t) I(t)
WL(t) = P(t) dt
=
VL(t) I(t) dt
= I(t) L dI(t) / dt dt
= L
I(t) dI(t)
WL(t) = L I 2
(t)
or
+
L
_
I(t)
I(t) VL(t)
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Example 5
Problem
Find the instantaneous energy in theinductor for the current shown in the figure
I (t) = 0 t < 1 ms
I (t) = 1/(4x10-3) (t-10-3) Amp 1 t 5 msI (t) = 1 Amp 5 t 9 ms
I (t) = 1/(4x10-3) (t - 13x 10-3) Amp 9 t 13 msI (t) = 0 t 13 ms I(t) (Amp)
0,0
0,5
1,0
0,0 2,0 4,0 6,0
t (msec)
8,0 10,0 12,0 14,0
V(t)
+
I(t)
+
L
L = 10 mH
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Example 5
Solution
By using the above formula
WL(t) = L I2(t)
W(t) = 0 Joule t < 1 ms
W(t) = 312.5 (t-10-3)2 Joules 1 t 5 ms
W(t) = 0.01 /2 x 12 = 0.005 Joules 5 t 9 msW(t) = 312.5 (t-13x10-3)2 Joules 9 t 13 msW(t) = 0 t 13 ms
W(t) (Joule)
0,001
0,002
0,003
0,004
0,005
0,006
0,0 2,0 4,0 6,0
t (msec)
8,0 10,0 12,0 14,0
V(t)
+
I(t)
+
L
L = 10 mH
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R-L Circuits
Problem
Solve the R-L circuit shown on the RHS
which consists of a resistance in serieswith an inductance for current waveformwhen the switch is turned on at time:t = 0 sec
A first order ordinary differential equation
V(t) = R I(t) + L dI(t) / dt
or
dI(t) / dt + (R/L) I(t) = (1/L) V(t)
Writing down KVL for the circuit
SolutionL
+
_
+
I(t)
V(t)
R
I(0)=I0
Switch
Switch is turned on at: t = 0 sec
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Solution of the resulting First Order Ordinary Differential Equation
Solve the resulting first order ordinary differential equation (ODE)
Solution
dI(t) / dt + (R/L) I(t) = (1/L) V(t)
dI(t) / dt + (R/L) I(t) = (1/L) V sin ( wt + )
V( t ) = V sin ( wt + )
^
100
0
Voltage (Volt)
Angle (Radians)
/2 3/2 2
200
300
312
Amplitude
Phase angle
^
L
+
_
+
I(t)
V(t)
R
I(0)=I0
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Solve the resulting first order ordinary differential equation (ODE)
Solution
(t) dI(t)/dt+ I(t) (1/RC)(t) = (t) ( V / L ) sin ( wt + )
(t) dI(t)/dt+ I(t) d/dt (t) = ( V / L ) (t) sin ( wt + )
d/dt [(t) I(t)] = ( V / L ) (t) sin ( wt + )
d/dt [(t) I(t)] dt = ( V / L ) (t) sin ( wt + ) dt + I(0)
(t) I(t) = ( V / L ) (t) sin ( wt + ) dt + I(0)
I(t) = I (t)-1
(t) sin (wt + ) dt + (t)-1
I(0)
^
^
^
^
^
^
Define an integration factor(t) = e t R/L
Multiply both sides of the above ODE by this factor;
^
100
0
Voltage (Volt)
Angle (Radians)
/2 3/2 2
200
300
312
Amplitude
Phase angle
dI(t) / dt + (R/L) I(t) = (V/L) sin ( wt + )^
Solution of the resulting First Order Ordinary Differential Equation
L
+
_
+
I(t)
V(t)
R
I(0)=I0
V( t ) = V sin ( wt + )
^
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Solution Continued)
Subsituting the integration factor(t) = e t R/L into
the above solution;
I(t) = I e t R/L
e t R/L sin (wt+ ) dt + e t R/L I(0)^
Switch is turned on at: t = 0 sec
Let = 0 (for simplicity)(R/L) sinwt - w coswt
e t R/L sin wt dt = e t R/L ----------------------------------
(R/L)2+ w 2
Taken from the book : Calculus and Analytic Geometry,
Thomas,Addison Wesley, Third Ed. 1965, pp. 369
Solution of the resulting First Order Ordinary Differential Equation
Switch
L
+
_
+
I(t)
V(t)
R
I(0)=I0
V( t ) = V sin ( wt + )^
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Solution of the resulting First Order Ordinary Differential Equation
Solution Continued)
Subsituting the above term into the solution;
Switch is turned on at: t = 0 sec
(R/L) sinwt - w coswtI(t) = I e t R/L e t R/L ---------------------------------- + e t R/L I(0)
(R/L)2+ w 2
^
(R/L) sinwt - w coswtI(t) = I ----------------------------------- + e t R/L I(0)
(R/L)2+ w 2
^
I
I(t) = -------------------(sinwt - (wL/R) coswt) + e t R/L I(0)(wL/R)2+ 1
^
Steady-State Term Transient Term
Switch
L
+
_
+
I(t)
V(t)
R
I(0)=I0
V( t ) = V sin ( wt + )^
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Numerical Example
Now assume that the parameters of the circuit on
the RHS are as follows;
V(t) = 312 sin wt Volts
R = 1 Ohms
L = 10 milli Henry
Switch
Switch is turned on at: t = 0 sec
I
I(t) = -------------------(sinwt - (wL/R) coswt) + e t R/L
I(0)(wL/R)2+ 1
^
Steady-State Term Transient Term
Solution of the resulting First Order Ordinary Differential Equation
L
+
_
+
I(t)
V(t)
R
I(0)=I0
V( t ) = V sin ( wt + )^
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Numerical Example
^
Steady-State Term Transient Term
Solution of the resulting First Order Ordinary Differential Equation
I
I(t) = -------------------(sinwt - (wL/R) coswt) + e t R/L I(0)(wL/R)2+ 1
^
-250
-200
-150
-100
-50
0
50
100
150
200
250
0.5 1 1.5 2 2.5 3 3.5 4
5
10
15
20
25
0 0.5 1 1.5 2 2.5 3 3.5 4
-300
-200
-100
0
100
200
300
400
0.5 1 1.5 2 2.5 3 3.5 4
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Solution for DC Voltage - Two Simple Rules
Rule - 1
An inductor with no initial current acts asan open circuit to a DC voltage sourceinitially
Then, the initial value of current will be;
I(0) = 0
The current waveform will then be;
I(t) = I() + [I(0) - I()] e-t/
I(0) = 0
= L / R = Time Constant: The time required foran inductor to reach 63 % of full current
= 1 x 2 mH = 1 x 0.002 = 0.002 Sec
L=0.002 H
OC
+ I(0) = 0
R
VDC
R = 1
+
_
+
I(t)
VDC0
I(0) = (1 / L)
V(t) dt = I0= 0 (OC)-
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Rule - 2
An inductor acts as a short circuit to a
DC current (or voltage) finally
Then, the final value of current will be;
I() = V / R
I() = V / R
R
+
_
+
I()VDC
+
_
+ I(t)
VDC
SC
V(t) = L d/dt I(t) = L d/dt (ct) = 0 (SC)
Solution for DC Voltage - Two Simple Rules
The current waveform will then be;
I(t) = I() + [I(0) - I()] e-t/
L=0.002 H
R = 1
= L / R = Time Constant: The time required foran inductor to reach 63 % of full current
= 1 x 2 mH = 1 x 0.002 = 0.002 Sec
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Definition of the Time Constant:
The current waveform will then be;
I(t) = I() + [I(0) - I()] e -t/
0
4.0
8.0
12.0
16.0
20.0
24.0
0 0,002 0,004 0,006 0,008 0,01
I(t)(Amps
)
t (sec)
% 63 of peak (24 V)
L = 2mH
+
_
+I(t)
V= 24 Volts
R = 1 Ohms
= L / R = Time Constant: The time required foran inductor to reach 63 % of full current= 1 x 2 mH = 1x 0.002 H = 0.002 Sec
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Example - 6
Problem
Find the current waveform in the 2 mHinductor shown on the RHS when it has aninitial current of 6 Amps and connected to a24 Volt DC voltage source through a wire with1 Ohm resistance
The current waveform will then be;
I(t) = I() + [I(0) - I()] e t /
I(t) = 24 + (6 - 24) e-t / 0.002= 24 - 18 e-t / 0.002 Amps
Inductor has 6 Amp initial current;
I(0) = I0= 6 Amp
Inductor will be SC at the end, hence;
I() = V / R = 24 / 1 = 24 Amps
0
4.0
8.0
12.0
16.0
20.0
24.0
0 0,002 0,004 0,006 0,008 0,01
I(t)(Amps)
t (sec)
I()
I(0)
L = 2mH
+
_
+I(t)
V= 24 Volts
R = 1 Ohms
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Example - 7
Problem
Solve the circuit shown on the RHS for
current waveform flowing in the inductor
+
_
L = 0.4 H+
I(t)V= 24 V
R1 = 10 Ohms
A
R2 = 5 Ohms
B
First take out the branch containing inductor,
and derive the Thevenin Equivalent of the LHScircuit seen from the terminals A and B
Solution
Kill the voltage source, and find Req.
R1 = 10 Ohms
+
V= 24 V R2 = 5 Ohms
AA
L = 0.4 H
B B
A
B
L = 0.4 H
R1 = 10 OhmsReq = 10 // 5
= 10 x 5 /(10 + 5)
= 10 / 3 Ohms
A
R2 = 5 Ohms
B
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Example 7 Continued)
Solution Continued)
Open circuit terminals A B and find VAB
Form the resulting Thevenin equivalentCircuit,
Connect the inductance to the resultingThevenin equivalent circuit,
Solve the resulting circuit by using thestraightforward method described inExample 6
R1 = 10 Ohms
+
V= 24 V
L = 0.4 H+
V= 24 / 3 = 8 V
R eq= 10 // 5 = 10 x 5 /(10+5)
= 10/3 Ohms
A
B
R2 = 5 Ohms
A
B
A
B
A
B
VA-B (t)
V = 24 VxR2/ ( R1 + R2)
= 24 x 5 / 15 = 24 / 3
= 8 V
L = 0.4 H
+
V= 24 / 3 = 8 V
R eq= 10 // 5 = 10 x 5 /(10+5)
= 10/3 Ohms
A
B
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R-L-C Circuits
Problem
Solve the following circuit which consists of a
resistance, an inductance and a capacitanceconnected in series for current waveform
V(t) = R I(t) + VL(t) + VC(t)
= R I(t) + L dI(t)/dt + (1/C)
I(t)dt + V(0)
Differentiating both sides wrt time once;
dV(t)/dt = R dI(t)/dt + Ld2I(t)/dt2+(1/C)I(t)
or
d2I(t)/dt2 + (R/L)dI(t)/dt + (1/LC) I(t) = (1/L) dV(t)/dt
A second order ordinary differential equation
Writing down KVL for the circuit;
Solution
V(t)
C+
_
+ I(t)
R
V(0)=V0
I(0)=I0
+
_
+
R L C
L
V(t)
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Differential Equation
VC(0) = VC0
IL(0) = IL0
d2I(t)/dt2 + (R/L)dI(t)/dt + (1/LC) I(t) = (1/L) dV(t)/dt
Initial Conditions
+
_
+
R L C
VC(0) = V(0) VL(0) VR(0)
= V(0) L d/dt IL(0) R IL(0)
The first initial condition may now be written as,
or
d/dt IL(0)= IL(0) =(1/L) [V(0) VC(0) R I L(0) ]
1
2Two initial conditions are needed for solution
V(t)C
+
_
+ V(0)=V0
I(t)
R I(0) = I0L
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R-L-C Circuit
d2I(t)/dt2 + (R/L)dI(t)/dt + (1/LC) I(t) = (1/L) dV(t)/dt
Solve the R-L-C circuit with the given parameters
shown on the RHS for current I(t)
d2I(t)/dt2 + (2/1) dI(t)/dt + 401 I(t) = (1/1) dV(t)/dt
d2I(t)/dt2+ 2 dI(t)/dt + 401 I(t) = d/dt (10 e-4t) = - 400 e-4t
+
C = 2.494 mF
+I(t)
R = 2 L = 1 H
V(t) = 100 e- 4 t
_
1 / (LC) = 401
Writing down KVL for the circuit shown on theRHS the following ODE is obtained
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+
_
+
R = 2
I(t)
L = 1 H
d2I(t)/dt2 + 2 dI(t)/dt + 401 I(t) = 0
First, obtain the homogeneous equation
by setting the RHS function to zero
Then, solve the characteristic equation
s2 + 2 s + 401 = 0
s1, s2= ( - b b2-4xaxc ) / (2 a)
= -1 j 20
Eigenvalues of the differential equation
V(t) = 100 e- 4 t
R-L-C Circuit
a =1 b = 2 c = 401
C = 2.494 mF
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V(t) = 100 e- 4 t
+
_
+
R = 2
I(t)
L = 1 H
Then, the solution becomes
e j= cos+ j sin = e - t[ k1 (cos 20 tj sin 20 t )= + k2(cos 20 t + j sin 20 t ) ]
I(t) = k1 es1 t + k2e
s2 t
= k1 e(-1 j20 ) t + k2e
(-1 + j20 ) t
= k1 e - t x e j20 t + k2 e - t x e j20 t
= e - t(k1 e j20 t + k2 e
j20 t) Eulers Identity
R-L-C Circuit
= 20 t
C = 2.494 mF
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Graphical Representaion
e j= cos+ j sin
Definition
^
1.0
cos
sin
e j = cos+ j sin = cos2+ sin2= 1
x
y
z
x2+ y2= z2
z = x2 + y2
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAOLU, Page 75
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Rearranging the solution terms
= e - t[ k1 (cos 20tj sin 20t )
= + k2(cos 20t + j sin 20t ) ]
= e
- t
[ (k1 + k2) cos 20t + (k2 k1) sin 20t ]
A B
R-L-C Circuit
I(t) = e - t( A cos 20t + B sin 20t)
Hence, the sinusoidal solution (decayingsinusoidal term) becomes;
V(t) = 100 e- 4 t
+
_
+
R = 2
I(t)
L = 1 H
Unknown coefficients to determined
C = 2.494 mF
METU
AC Circuits
Solution
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Nonhomogeneous Term
Transient Term)
Now the nonhomogeneous (transient) term in thesolution is to be determined V(t) = 100 e- 4 t
+
_
R = 2
+
I(t)
L = 1 H
Definition of the Nonhomogeneous Term
General form of the nonhomogeneous (transient)term in the solution may be expressed as
In(t) = c e 4 t
where, In(t) is the nonhomogeneous term in the solution,c is an unknown coefficient to be determined
C = 2.494 mF
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AC Circuits
Solution
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Substitute the nonhomogeneous solution;
In(t) = c e 4 t
to the given differential equation;
and solve it for the unknown coefficient c
d2I(t)/dt2 + 2 dI(t)/dt + 401 I(t) = - 400 e-4t
Nonhomogeneous Term
Transient Term)
d2(ce 4 t)/dt2 + 2 d(ce 4 t)/dt + 401 ce 4 t = -400 e-4t
16c e 4 t+ 2c(-4 e 4 t ) + 401 c e 4 t = - 40 e-4 t
V(t) = 100 e- 4 t
+
_
+
R = 2
I(t)
L = 1 H
C = 2.494 mF
These terms cancel
METU
AC Circuits
Solution
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAOLU, Page 79
Transient Term
16 c - 8 c + 401 c = - 400
409 c = - 400 or c = - 400 / 409 = - 0.97799
Transient term then becomes;
In
(t) = - 0.97799 e 4 t
d2(ce 4 t)/dt2 + 2 d(ce 4 t)/dt + 401 ce 4 t = -400 e-4t
16c e 4 t+ 2c(-4 e 4 t ) + 401 c e 4 t = - 40 e-4 t
These terms cancel
V(t) = 100 e- 4 t
+
_
+I(t)
R = 2 L = 1 H
C = 2.494 mF
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METU
AC Circuits
Solution
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAOLU, Page 81
V(t) = 100 e- 4 t
+
_
+
R = 2
I(t)
L = 1 H
I(t) = - 0.97799 e 4 t
+ e- t
( A cos 2t + B sin 2t)
Transient Term
Determination of the Unknown
Coefficients
The above solution must satisfy the giveninitial conditions;
Decaying sinosoidal Term
C = 2.494 mF
IL (0) = IL0= 32.5 A
VC(0) = VC0 = 22 Volts
METU
AC Circuits
Solution
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Determination of the Unknown
Coefficients
IL (0) = IL0= 32.5 A
VC(0) = VC0 = 22 Volts
Substitute the given initial conditions intothe complete solution equation and solve forthe unknown coefficients A and B;
V(t) = 100 e- 4 t
+
_
+
R = 2
I(t)
L = 1 HI(t) = - 0.97799 e 4 t
+ e- t
( A cos 2t + B sin 2t)
Transient Term Decaying sinosoidal TermC = 2.494 mF
METU
AC Circuits
Solution
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAOLU, Page 83
Determination of the Unknown
Coefficients
Substitute the given initial conditions into thecomplete solution equation and solve for theunknown coefficients A and B;
A = 0.97799
IL(0) = - 0.97799 e0 + e 0( A cos 0 + B sin 0)
= - 0.97799 + A = 0
IL(t) = - 0.97799 e 4 t + e - t( A cos 20t + B sin 20t)
V(t) = 100 e- 4 t
+
_
+
I L(t)
R = 2 L = 1H
C = 2.494 mF
IL (0) = IL0= 32.5 A
VC(0) = VC0 = 22 Volts
METU
AC Circuits
Solution
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAOLU, Page 84
V(0) = 100 e- 4 t
= 100 e- 4 x 0
= 1000 0
Determination of the Unknown
Coefficients
IL (0) = IL0= 32.5 A
VC(0) = VC0 = 22 Volts
IL(t) = - 0.97799 e 4 t
+ e- t
( A cos 20t + B sin 20t)
d/dt IL(0) = IL(0) = (1/L) [ V(0) VC(0) R IL(0) ]
= (1/1) (100 22 2 * 32.5)= 13 Amp/sec
V(t) = 100 e- 4 t
+
_
+
I L(t)
R = 2 L = 1 H
C = 2.494 mF
METU
AC Circuits
Solution
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Determination of the Unknown
Coefficients
d/dt IL(t) = 0.97799 x 4 e 4 t
- e- t
(A cos 20t +B sin 20t)+ e - t( -20 A sin 20 t + 20 B cos 20t )
d/dt IL(0) = 0.97799 x 4 e0- e 0(A cos 0 + B sin 0)
+ e 0(-20 A sin 0 + 20 B cos 0 ) = 13
= 0.97799 x 4 A + 20B = 13
= 0.97799 x 4 0.97799 + 20B = 13
B = (13 - 3 x 0.9799) / 20 = 0.5033
V(t) = 100 e- 4 t
+
_
+
I L(t)
R = 2 L = 1 H
C = 2.494 mF
METU
AC Circuits
Solution Terms
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IL(t) = - 0.9799 e 4 t + e - t( 0.97799 cos 20t + 0.5033 sin 20t)
General form of the Solution
Decaying sinosoidal TermTransient Term
+
_
V(t) = 100 e- 4 t
+
I(t)
R = 2 L = 1 H
C = 2.494 mF
METU
AC Circuits
Solution Terms
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Transient Term
Transient Term
+
_
V(t) = 100 e- 4 t
+
I(t)
R = 2 L = 1 H
C = 2.494 mF
IL(t) = - 0.9799 e 4 t + e - t( 0.97799 cos 20t + 0.5033 sin 20t)
-1,20
-1,00
-0,80
-0,60
-0,40
-0,20
0,000,0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,6
METU
AC Circuits
Solution Terms
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAOLU, Page 88
Sinosoidal Terms
Sinusoidal Terms
0.9779 cos 20t + 0.5033sin 20t
0.9779 cos 20t
0.5033 sin 20t
+
_
V(t) = 100 e- 4 t
+
I(t)
R = 2 L = 1 H
C = 2.494 mF
IL(t) = - 0.9799 e 4 t + e - t( 0.97799 cos 20t + 0.5033 sin 20t )
-1,20
-0,80
-0,40
0,00
0,40
0,80
1,20
0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1
METU
AC Circuits
Solution Terms
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAOLU, Page 89
Exponentially Decaying Sinusoidal Term
V(t) = 100 e- 4 t
+
_
+
I(t)
R = 2 L = 1 H
C = 2.494 mF
IL(t) = - 0.9799 e 4 t + e - t( 0.97799 cos 20t + 0.5033 sin 20t )
-1,20
-0,80
-0,40
0
0,40
0,80
1,20
0,5 1 1,5 2 2,5 3 3,5 4
e - t
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AC Circuits
Solution Terms
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Overall Solution
V(t) = 100 e- 4 t
+
_
+
I(t)
R = 2 L = 1 H
C = 2.494 mF
IL(t) = - 0.9799 e 4 t + e - t( 0.97799 cos 20t + 0.5033 sin 20t )
-1,60
-1,20
-0,80
-0,40
0,0
0,40
0,80
0,5 1,0 1,5 2,0 2,5 3,0 3,5 4,0
METU
AC Circuits
The Concept of RMS Value
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAOLU, Page 91
Definition: RMS Root Mean Square)
Justification is based on the principle ofequating the currents calculated in both cases.
First, calculate the power dissipated (lost asheat) in the resistance R in the DC circuitshown on the RHS
VR = R xIDCP = VRxIDC
= R xIDC2
VDC
_
+
IDC
VR
RMS value of an AC current is the value of DCcurrent that would dissipate the same amountof power as the AC current on a resistance R
Justification
+
R
METU
AC Circuits
The Concept of RMS Value
I 2(t) (I i t )2I(t) I i t
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAOLU, Page 92
Now, calculate the power dissipated (lostas heat) in the resistance R in the AC
circuit shown on the RHS
V (t) = I(t) xR
P (t) = V(t) xI(t)
= R I(t)2
= R (Vmax/ R sin wt )2
= R (I maxsin wt )2
R
V(t) = Vmaxsin wt
_
+
I(t)
V(t)
+
V(t)
I(t) = (Vmax/ R) sin wt
Imax
-2,0
-1,5
-1,0
-0,5
0
0,5
1,0
1,5
2,0
2,5
Angle (Radians)
/2 3/2 20
I 2(t) = (Imaxsin wt )2I(t) = Imaxsin wtMS Root Mean Square)
METU
AC Circuits
The Concept of RMS Value
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAOLU, Page 93
Problem
Let us now calculate the average of P(t)
Pavg = (1/T) P(t) dt
= (1/T)
R I(t)2 dt
= R (1/T)
I(t)2dt
= R Irms2
where I(t)rms = ((1/T) I(t)2dt)1/2
I(t)rms
=
( 1/T ) I(t)2dt
0
0,5
1,0
1,5
2,0
2,5
Angle (Radians)
/2 3/2 20
Irms = ( 1/T ) I(t)2dtI2(t) = ( Imaxsin wt )2
METU
AC Circuits
Definition of RMS Value
I2(t) ( I i t )2
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Problem
RMS value of an AC current is the valueof DC current that would dissipate the
same amount of power as AC current ona resistance R
_
+
IDC
R
+
VR
_
+
I(t)
V(t) +
VR(t)
Pavg= R IDC2 Pavg= R I(t)rms
2=
VDC
R
I2(t) = ( Imaxsin wt )2
0
0,5
1,0
1,5
2,0
2,5
Angle (Radians)
/2 3/2 20
Irms = ( 1/T ) I(t)2dt
METU
AC Circuits
RMS Value of a Sinusoidal Waveform
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAOLU, Page 95
RMS value of a sinusoidal waveform maybe expressed as 0.7071 multiple of thepeak value of the waveform as follows;
I (t) = I maxsin wt
Irms = ((1/T) I(t)2dt)1/2
= ((1/T) (I maxsin wt)2dt)1/2
= ((1/T) (Imax
2 sin2w td t )1/2
= ((1/T) Imax2 (1 cos2wt) dt )1/2
= ((1/T) Imax2 ( d t cos2wt dt ))1/2
= ((1/T) Imax2( T / 2 x0 ))1/2
= ((1/T) T Imax2/ 2 )1/2
= ( Imax2/ 2 )1/2
= Imax/ 2 = I maxx0.7071-2,0
-1,5
-1,0
-0,5
0
0,5
1,0
1,5
2,0
Current (Amp)
Angle (Radians)
/2 3
/22
sin2wt = 1 cos2wt
= 1 (1 + cos 2wt) / 2
= 1 cos 2wt
= cos2wt
METU
AC Circuits
RMS Value of a Sinusoidal Waveform
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAOLU, Page 96
Calculate the RMS value of thesinusoidal voltage waveform shown onthe RHS
Vrms = Vmax/ 21/2 = V maxx0.7071
= 312 x 0,7071 = 220 Volts
Problem
Domestic rms voltage level in Turkey
312
- 312
0
V
oltage(Volts)
Angle (Radians)
/2 3/2 2
Vrms = 220 Volts
220
METU
AC Circuits
Example - 8
V = ( 1/T ) V(t)2 dt
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Problem
Calculate the RMS value of a the voltagewaveform shown on the RHS
Vrms = ( 1/T ) V(t)2dt
-4
-2
0
2
4
0,1 0,2 0,3 0,4 0,5 0,6
Time (msec)Voltage(Volts)
Voltage2(Volts)
0
8
16
24
32
0,1 0,2 0,3 0,4 0,5 0,6
Time (msec)
V2(t) = 16
_
+
I(t)
V(t) +
VR(t)
Vrms = ( 1/T ) V(t)2dt
Vrms = ( 1/0.3) ( 42dt + (-4)2dt)0
0.1
0.1
0.3
= 4 Volts
METU
AC Circuits
Any Questions Please ?
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