ee-221-review of dc circuits
TRANSCRIPT
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Ohms Law
Ohms law states that the voltage acrossa resistor is directly proportional to the
current I flowing through the resistor.
Mathematical expression for Ohms Law :
Note: the current enters the positive
side and leaves the negative side of v.
Two extreme possible values of R: R = 0 (zero) v = 0 V -----short circuit
R =
(infinity) I = 0 A --- open circuit
2
iRv
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Ohms Law
Conductance is the ability of an element to conductelectric current; it is the reciprocal of resistance R and ismeasured in Siemens (S).
The power dissipated by a resistor:
R
vRivip
2
2
3
v
i
RG 1
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Nodes and Branches
A branchrepresents a single element such as a voltage sourceor a resistor (or a series connection of elements).
A nodeis the point of connection between two or more
branches.
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Kirchhoffs Current Law (KCL)
Kirchhoffs current law (KCL) states that the algebraicsum of currents entering a node (or a closedboundary) is zero.
5
01
N
n
niMathematically,
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Kirchhoffs Voltage Law (KVL)
Kirchhoffs voltage law (KVL) states that the algebraicsum of all voltages around a closed path (or loop) is
zero.
6
Mathematically, 0
1
M
m
nv
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Series Resistors and voltage Division
The equivalent resistance of any number of resistors
connected in a series is the sum of the individual
resistances.
The voltage divider can be expressed as
7
N
n
nNeq RRRRR1
21
vRRR
Rv
N
n
n
21
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Parallel Resistors and Current Division
The equivalent resistance of a circuit with N resistors inparallel is:
The total current i is shared by the resistors in inverseproportion to their resistances. The current divider can be
expressed as:
8
Neq RRRR
1111
21
n
eq
n
n
R
iR
R
vi
http://en.wikipedia.org/wiki/File:Resistors_in_parallel.svg -
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Superposition Theorem
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Example 1: Find v in the circuit shown below.
3A is discardedby open-circuit
6V is discardedby short-circuit
Answer: v= 10V
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Source Transformation
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Source transformation is the process of replacing avoltage source vSin series with a resistor Rby anequivalent circuit that consists of a current sourceiSin parallel with a resistor R,or vice versa.
R
vi
Riv
s
s
ss
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Source Transformation
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Example 1: Find ioin the circuit shown below.
Answer: io= 1.78A
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TheveninsTheorem
13
A linear two-terminalcircuit can be replaced byan equivalent circuitconsisting of a voltagesource VThin series witha resistor RTh,
where
VThis the open-circuitvoltage across terminals a-b.
RThis the equivalentresistance of the linearcircuit with the independentsources turned off.
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TheveninsTheorem
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Example 1: Find the Thevenins equivalent circuit to theleft of the terminals in the circuit shown below.
Answer: VTh= 6V, RTh= 3W, i = 1.5A
6
4
(a)
RTh
6
2A
6
4
(b)
6 2A
+
VTh
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Maximum Power Transfer
L
Th
ThL
RVPRR4
2
max
L
LTh
Th
L R
RR
VRiP
2
2
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The power consumed by the loadresistance RL is given by
Power transfer profile with differentvalues of RL
Maximum power is obtainedby dP/dRL= 0.
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Capacitors
A capacitor is a passive element designed to store energy
in its electric field. It consists of two conducting plates
separated by an insulator (or dielectric).
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Capacitors
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Current-voltage relationship ofcapacitor according to aboveconvention:
td
vdCi )(
10
0
tvtdiC
vt
t
and
Energy stored in a capacitor:
2
2
1vCw
A capacitor acts as opencircuitunder constant voltage.
The voltage across a capacitorcannot change abruptly
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Capacitors
Example 1: The current flow into an initially discharged 1mF capacitoris shown below. Calculate the voltage across its terminals at t= 2 msand t= 5 ms.
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Answer: v(2ms) = 100 mV, v(5ms) = 500 mV
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Series and Parallel Capacitors
The equivalent capacitance of Nparallel-connected
capacitors is the sum of the individual capacitances.
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Neq CCCC ...21
The equivalent capacitance of Nseries-connectedcapacitors is the reciprocal of the sum of the reciprocals ofthe individual capacitances.
Neq CCCC
1...
111
21
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Series and Parallel CapacitorsExample 2:Find the equivalent capacitance seen at the
terminals of the circuit in the circuit shown below:
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Answer: Ceq= 40F
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Inductors
An inductor is a passive element designed to store
energy in its magnetic field. It consists of a coil of
conducting wire.
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InductorsExample 1: Determine vc, iL, and the energy stored in the
capacitor and inductor in the circuit of circuit shown below
under steady-state conditions.
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Answer: iL= 3A, vC= 3V,WL= 1.125J, WC= 9J
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Series and Parallel Inductors The equivalent inductance of series-connected
inductors is the sum of the individual inductances.
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Neq LLLL ...21
The equivalent capacitance of parallelinductors is the
reciprocal of the sum of the reciprocals of the individual
inductances.
Neq LLLL
1...
111
21
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RC Circuit (source free)
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Currentflow inresistor R
Current flow incapacitor C
0dt
dvCR
v0CR
ii
By KCL
Applying Kirchhoffs laws to purely resistive circuit
results inalgebraic equations. Applying the laws to RC and RL circuitsproduces first-
first-order differential equations.
Vois the initial voltage= RC is the time constant
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RC Circuit (source free)
Example 1: Refer to the circuit below, determine vC, and io
for t 0. Assume that vC(0) = 30 V.
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Answer: vC = 30e0.25tV;
io = 2.5e0.25tA
Example 2: The switch in circuit below was closed for a long
time, then it opened at t = 0, find v(t) for t 0.
Answer: V(t) = 8e2tV
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RL Circuit (source free)
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0RL
vvBy KVL
0 iRdt
di
L
0)( iL
R
dt
di /0)(
teIti
Iois the initial current= L/R is the time constant
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RL Circuit (source free)
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Example 2: For the circuit, find i(t) for t > 0.
Answer: i(t) = 2e2tA
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Unit-Step Function
The unit step functionu(t) is 0 for negative values of
t,and 1 for positive values of t.
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0,1
0,0)(
t
t
tu
o
o
o
tt
tt
ttu
,1
,0)(
o
o
o
tt
tt
ttu
,1
,0)(
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Unit-Step Function
1. voltage source.
2. for current source:
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Represent an abrupt change:
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Step-Response of an RC CircuitThe step response of a circuit is its behavior when theexcitation is the step function, which may be a voltage or a
current source.
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Initial voltage (given):v(0-) = v(0+) = V0
Applying KCL,
or
Where u(t) is the unit-step function
0)(
R
tuVv
dt
dvc
s
RC
tuVv
RCdt
dvs
)()
1(
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The Step-Response of a RC Circuit
Integrating both sides and considering the initialconditions, the solution of the equation is:
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0)(0)(
/
0
0
teVVV
tVtv
t
ss
Final valueat t ->
Initial valueat t = 0
Source-freeResponse
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Step-Response of a RC Circuit
Example 1: Find v(t) for t > 0 in the circuit in below.
Assume the switch has been open for a long time and isclosed at t = 0. Calculate v(t) at t = 0.5.
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Answer: and v(0.5) = 0.52 V515)( 2 tetv
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Step-response of a RL Circuit
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Initial current (given)
i(0-) = i(0+) = Io
Final inductor currenti() = Vs/R
Apply KVL:
Time constant = L/R
0)()(
teR
VI
R
Vti
t
so
s
L
tuVi
L
R
dt
dis
)()(
/
)]()0([)()(
teiiiti
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Second-Order Circuits
Chapter 8
8.1 Examples of 2nd order RCL circuit
8.2 The source-free series RLC circuit8.3 The source-free parallel RLC circuit
8.4 Step response of a series RLC circuit
8.5 Step response of a parallel RLC
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Examples of 2nd Order RLC circuits
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A second-order circuit is characterized by asecond-order differential equation.It consists ofresistorsand the equivalent of two energy
storage elements.
RLC Series RLC Parallel RL T-config RC Pi-config
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Source-Free Series RLC Circuits
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The solution of the source-freeseries RLC circuit is called as thenatural responseof the circuit.
The circuit is excitedby the energyinitially stored in the capacitor andinductor.
02
2
LC
i
dt
di
L
R
dt
id
How to derive and how to solve?
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02 2
02
2
idt
di
dt
id
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There are three possible solutions for the following 2ndorder differential equation (where the constants A1,
A2, B1, B2 are found from the initial conditions.
1. If > o, over-damped casetsts
eAeAti 21 21)( 2
0
2
2,1 swhere
2. If = o
, critical damped case
tetAAti
)()( 12 2,1swhere
3. If < o, under-damped case
)sincos()( 21 tBtBeti ddt
where22
0
d
Source-Free Series RLC Circuits
LCand
L
R 1
20 where
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Source-Free Series RLC Circuits
Example 1
If R = 10 , L = 5 H, and
C = 2 mF, find and 0.
What type of natural
response will the circuithave?
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Answer: = 1, 0 = 10, under-damped
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Source-Free Series RLC Circuits
Example 2The circuit shown below
has reached steady state
at t = 0-. If the make-
before-break switch
moves to position b at t =
0, calculate i(t) for t > 0.
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Answer: i(t) = e2.5t[5cos1.66t 7.538sin1.66t] A
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Source-Free Parallel RLC Circuits
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011
2
2
vLCdt
dv
RCdt
vd
0
0 )(1
)0( dttvL
IiLet
v(0) = V0
Apply KCL to the top node:
t
dt
dvCvdt
LR
v0
1
Taking the derivative with
respect to t and dividing by C
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Source-Free Parallel RLC Circuits
LCRCv
dt
dv
dt
vd 1and
2
1where02
0
2
02
2
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There are three possible solutions for the following2nd order differential equation:
1. If > o, over-damped casetsts
eAeAtv 21 )( 21 2
0
2
2,1 swhere
2. If = o
, critical damped case
tetAAtv
)()( 12 2,1swhere
3. If < o, under-damped case
)sincos()( 21 tBtBetv ddt
where22
0
d
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Step-Response Series RLC Circuits
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The step responseis obtained by thesudden applicationof a dc source.
LC
v
LC
v
dt
dv
L
R
dt
vd s
2
2
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Step-Response Series RLC Circuits
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The solution of the equation should have two components:the transient response vt(t)& the steady-state response vss(t):
)()()( tvtvtvsst
The transient response vtis the same as that for source-free case
The steady-state response is the final value of v(t).
vss(t) = v()
The values of A1and A2are obtained from the initial conditions:
v(0) and dv(0)/dt.
tsts
t eAeAtv 21 21)( (over-damped)
t
t etAAtv
)()( 21 (critically damped)
)sincos()( 21 tAtAetv
dd
t
t
(under-damped)
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Step-Response Series RLC Circuits
Example 4Having been in position for a long time, the switch in thecircuit below is moved to position b at t = 0. Find v(t)and vR(t) for t > 0.
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Answer: v(t) = 10 +e-2t(2cos3.46t 1.15sin3.46t) V
vR(t)= e2t
[2.31sin3.46t] V
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Step-Response Parallel RLC Circuits
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LC
I
LC
i
dt
di
RCdt
ids
1
2
2
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8.5 Step-Response Parallel RLC Circuits
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The solution of the equation should have two components:the transient response vt(t)& the steady-state response vss(t):
)()()( tititisst
The transient response itis the same as that for source-free case
The steady-state response is the final value of i(t).
iss(t) = i() = Is
The values of A1and A2are obtained from the initial conditions:
i(0) and di(0)/dt.
tsts
t eAeAti 21 21)( (over-damped)
t
t etAAti
)()( 21 (critical damped)
)sincos()(21
tAtAetidd
t
t
(under-damped)
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Step-Response Parallel RLC Circuits
Find i(t) and v(t) for t > 0 in the circuit shown in circuit
shown below:
Answer: v(t) = Ldi/dt = 5x20sint = 100sint V