Circuitry Ch07 First-Order Circuits
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Transcript of Circuitry Ch07 First-Order Circuits
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10
Chapter 7 First-Order Circuits
Circuit elements Passive elements : resistors, capacitors and inductors Active element : OP Amp (operational amplifier)
Analysis of RC and RL circuits by applying Kirchhoffs laws RC and RL circuits are first-order circuit A first-order circuit is characterized by a first-order
differential equation Excitation of first-order circuits Source-free circuit : initial conditions on inductor and capacitor Independent sources : dc, sinusoidal and exponential sources
1
7.2 The Source-Free RC Circuit
A source-free RC circuit occurs when dc source is suddenly removed
The resistor and capacitor may be equivalent circuit of combinations of them
We want to know the circuit response, i.e. the capacitor voltage v(t). Assuming, v(0) = V0
The energy stored is
Applying KCL:
20CV2
10w =)(
0RCv
dtdv0
Rv
dtdvC
0ii RC
=+=+=+
or
-
22
The Source-Free RC Circuit
This is a first-order differential equation
Integrating both sidesdt
RC1
vdv =
RCt0
0
RCt
eVtvVA0v
AetvRCt
Avln
AlnRCtv ln
/
/
)(
)( ,conditions initial theFrom)(
constant nintegratio theis A where
===
==
+=
An exponential decay of the initial voltage
3
The Source-Free RC Circuit
The natural response of a circuit refers to the behavior (in terms of voltages and currents) of the circuit itself, with no external sources of excitation
Natural response depends only on the nature of the circuit alone, nothing related to the external sources
The time constant ( = RC)of a circuit is the time required for the response to decay by a factor of 1/e or 36.8% of its initial value
/
/
)(
.)(
t0
01
0RC
0
eVtvand
V3680eVeVv
=
===
-
34
Steady state v.s. transient state
v(t) is less than 1% of V0 after 5; in other words,It takes 5 for the circuit to reach its final state or steady statewhen no changes occur
The state before steady state is called transient state The smaller the time constant, the faster the response
0.0067450.0183240.0497930.1353420.36788v(t)/V0t
5
Energy in RC circuit
)( toequal is which,)( , tas that Notice
ere wh)(
)(
is time toupresistor by the absorbedenergy The
)(
isresistor thein dissipatedpower The
)()(current resistor The
/
//
/
/
0wCV21w
RCe1CV21
0t
eR2
VdteR
Vpdttw
t
eR
Vvitp
eRV
Rtvti
C2
0R
t220
t22
0t
0
t22
0t
0R
t22
0R
t0R
==
===
==
==
-
46
Another perspective of time constant
The time constant is the initial rate of decay, or the time taken for v/V0 to decay from unity to zero, assuming constant rate of decay
This initial slope interpretation is used to find from the response curve
Draw the tangent to the curve at t = 0 The tangent intercepts with the time axis at t =
1e1
Vv
dtd
0t
t
0t0
==
=
=
/ /)( t0eVtv=
7
Work with a Source-free RC Circuit
First find: The initial voltage v(0) = V0 across the capacitor The time constant
The capacitor voltage is then
The time constant is identical regardless of what output is When a circuit contains a single capacitor and several resistors
or dependent sources, the Thevenin equivalent can be found at the terminals of the capacitor
/)()()( tC e0vtvtv==
-
58
Example 7.1
In the figure, let vC(0) = 15V. FindvC, vx, and ix, for t > 0.
Sol: The equivalent resistance
The time constant Thus
20 5 420 5eq
R = = +4(0.1) 0.4eqR C s = = =
/ / 0.4
2.5
2.5 2.5
2.5
(0) 15 ,15
12 0.6(15 ) 912 8
0.7512
t t
tc
t tx
txx
v v e e Vv v e V
v v e e V
vi e A
= == == = =+= =
9
Practice Problem 7.1
For the circuit, let vC(0) = 30V. Determine vC, vx, and io, for t 0.
Ans: 30e-0.25tV, 10e-0.25tV, -2.5e-0.25tV
-
610
Example 7.2
The switch in the circuit has beenclosed for a long time, and it is opened at t = 0. Find v(t) for t 0 and the initial energy stored in the capacitor.
Sol: Using voltage division
vC(0) is the same as vC(0-)
9( ) (20) 15 , 09 3c
v t V t= =
-
712
7.3 The Source-Free RL Circuit
Analysis goal is to find the circuit response Select inductor current as the response in order
to take advantage of that: inductor current cannot change instantaneously
Assuming i(0) = I0, the energy stored in the inductor is
This is a first-order differential equation
0iLR
dtid0Ri
dtidL
0vv RL
=+=+=+
or
: KVLApplying
20LI2
10w =)(
13
The Source-Free RL Circuit
Rearranging
Integrating both sidesdt
LR
iid =
where)(
)(
)( )(
//
)(
RLeIeIti
LRt
Itiln
0LRtIlntiln
0t
LR
Iti
iln
dtLR
iid
t0
LRt0
0
00
t
0
ti
I0
===
=
+==
=
An exponential decay of the initial current
-
814
Natural response of RL circuit
The natural response of the RL circuit is an exponential decay of the initial current
/0
2 2 /0
2 2 /00 0
2 2 / 2 2 /0 0
0
20
( )
( )
1 1 (1 )2 2
1Note that as , ( ) , which is equal to (0)2
tR
tR
t t tR
tt t
R L
v t iR I Rep v i I Re
w t pdt I Re dt
I Re LI e
t w LI w
= == =
= =
= =
15
Key to Work with a Source-free RL Circuit
First find: The initial current i(0) = I0 through the inductor The time constant of the circuit
The inductor current is then
Once iL is determined, inductor voltage vL, resistor voltage vRand resistor current iR can be obtained
When a circuit contains a single inductor and several resistors or dependent sources, the Thevenin equivalent can be found at the terminals of the inductor
/)()()( tL e0ititi==
-
916
Example 7.3
Assuming that i(0)=10A, calculate i(t) and ix(t) in the circuit.
Sol: Method 1:
1 2 1 2
2 1 1 2 1
1 0 1
0
0
12( ) 1 02
56 2 3 06
3 , 31Hence 3eq Th
Applying KVL
i i i i
i i i i i
i A i i AvR Ri
+ = =
= == = =
= = =
17
Example 7.3
Method 2:/ (2 / 3)
3The timeconstant is 2
Thus, thecurrent through the inductor is( ) (0) 10 0
eq
t t
L sR
i t i e e A t
= =
= = >
11 2
2 1 1 2 1
11
1
1
11 2( ) 02
52 6 2 3 06
2 03
2Rearranging terms3
diApply KVL for loop i idt
for loop i i i i i
diSubstituting idt
di dti
+ =
= =
+ =
=
-
10
18
Example 7.3
Practice Problem 7.3 Let i(0)=5A, find i(t) and vx(t) in the
circuit.
Ans: 5e-53t A, -15e-53t V
( )
(0)0
(2 /3)
2ln3
( ) (0)
ti t
ii t
i t i e
= = (2 /3)
(2 /3) (2 /3)
(2 /3)
10 02 100.5(10)( )3 3
( ) 1.667 , 02
t t
t t
tx
e A tdiv L e e Vdtvi t e A t
= >= = =
= = >
19
Example 7.4
The switch in the circuit has been closed for a long time. At t = 0, the switch is opened. Calculate i(t) for t > 0.
Sol:
1
1
Combine the 4 and 124 12resistors 34 12
40Hence, 82 3
Usingcurrent division12( ) 6 , 0
12 4(0) (0 ) 6
i A
i t i A t
i i A
= += =+
= =
-
11
20
Example 7.4
Combining the resistors, we have
The time constant is
Thus,
Practice Problem 7.4 For the circuit, find i(t) for t > 0.
Ans: 2e-2t A
816||)412( =+=eqRs
41
82 ===
eqRL
A6)0()( 4/ tt eeiti ==
21
Example 7.5
In the circuit, find io, vo, and i for all time, assuming that the switch was open for a long time.
Sol: For t < 0
For t > 0A)(
V)()( A,)(
20i
6ti3tv232
10ti 0
====+=
A)()(
s
||
/ ttTh
Th
e2e0iti
1RL
263R
======
-
12
22
Example 7.5
Since the inductor is in the parallel with the 6 and 3
0
0
0 0
( ) 2( 2 ) 4 , 0
2( ) , 06 3
0 , 0 6 , 0( ) , ( )2 , 0 4 , 0
32 , 0
( )2 , 0
t tL
tL
t t
t
div t v L e e V tdt
vi t e A t
ThusA t V t
i t v te A t e V t
A ti t
e A t
= = = = >
= = >
-
13
24
7.4 Singularity Functions ()
Singularity functions are functions that either are discountinuous or have discountinuous derivatives.
Singularity functions (also called switching functions) serve good approximations to the switching signals in the circuits with switching operations
These functions are used to describe the sudden application of an independent dc voltage or current source
Three most widely used are unit step, unit impulse, and unit ramp functions
25
Unit step functions
The unit step function u(t) is 0 for negative values of t and 1 for positive values of t.
>
-
14
26
Unit step voltage source
Step function is used to represent an abrupt change in voltage or current, like the changes that occur in the circuits of control system or digital computers
)()( aslly mathmatica drepresente be canIt ,,
)( like is a voltage If
00
00
0
ttuVtvttVtt0
tv
=
>
-
15
28
Unit impulse function
Unit impulse function (t) is zero everywhere except at t = 0, where it is undefined
Unit impulse function is the derivative of the unit step function
Unit impulse function is also known as delta function The unit area is known as the strength
of the impulse function
0, 0( ) ( ) , 0
0, 0
tdt u t Undefined tdt
t
unit area
0
0( ) ( ) 1t dt t dt + = =
29
If an impulse function has a strength other than unity, the area of the impulse is equal to its strength
Integrating the function results in the value of the function exactly at the point where the impulse occurs
Useful feature known as sampling or sifting property
Sampling or sifting property
0
0
0
0
0 0
0 0
0 0 0
( ) ( ) where
( ) ( )
( ) ( ) ( )
b
a
t
t
t
t
f t t t dt a t b
f t t t dt
f t t t dt f t
+
+
<
=+=+
or
For
)(or )(
0
( )
0
00
ln( )
ln( ( ) ) ln( ) 0 or ln
tv t
s V
ss s
s
tv VRC
t v V tv t V V VRC V V RC
= = + =
-
23
44
Step Response of an RC Circuit
Taking the exponential of both sides
Thus,
This is known as the complete response (or total response) of the RC circuit
0/
0
, 0( )
( ) , 0ts s
V tv t
V V V e t
assuming Vs > V0
45
Step Response of an RC Circuit
If assuming the capacitor is initially uncharged, then V0 = 0
The above can be alternatively written as
The capacitor current is obtained by i(t) = Cdv/dt
/
0, 0( )
(1 ), 0ts
tv t
V e t
/( ) (1 ) ( )tsv t V e u t=
/
/
( ) , 0
or ( ) ( )
ts
ts
dv Ci t C V e tdtVi t e u tR
= = >
=
-
24
46
Complete response
Complete response = nature response + forced response
Originally Alternatively
Natural response vn is the source-free response, dying out gradually with the transient part of the forced response
Forced response vf is the response induced by external force, existing continually to form the steady-state component
stored energy independent source
)( , where // tsft
0n
fn
e1VveVv
vvv ==
+=0( ) ( ) , 0
t
s sv t V V V e t= + >
0
V0
Vs
vn
vfvtotal
47
Complete response
Complete response = transient response + steady-state response
The transient response is the circuits temporary response that will die out with time
The steady-state response is the behavior of the circuit a long time after an external excitation is applied
Temporary part Permanent part
ssst
s0t
sst
VveVVvvvv
==+=
, )( where /
0
0
s
vt
vssvtotal
V
V
-
25
48
Decomposition of complete response
The first decomposition of the complete response is in terms of the source of the responses; the latter is the permanency of the responses
Under certain conditions, the natural and transient responses are the same; so as forced and steady-state responses
Either way, complete response can be written as
where v(0) is the initial voltage at t = 0+ and v() is the final or steady-state value
/( ) ( ) [ (0) ( )] tv t v v v e = +
49
Solving RC circuit with response decomposition
To find the step response of an RC circuit requires three things The initial capacitor voltage v(0) The final capacitor voltage v() The time constant
If the switch changes position at time t = t0, there is a time delay in the response
where v(t0) is the initial value at t = t0+
0( ) /0( ) ( ) [ ( ) ( )]
t tv t v v t v e = +
-
26
50
Example 7.10
The switch in the figure has been in position A for a long time. At t = 0, the switch moves to B. Determine v(t) for t > 0 and calculate its value at t = 1s and 4s.
Sol: By voltage division
3 3
5(0 ) (24) 155 3
Since the capacitor voltage cannot change(0) (0 ) (0 ) 15
The time constant 4 10 0.5 10 2Th
v V
v v v VR C s
+
= =+
= = == = =
51
Example 7.10
Since the capacitor acts like an open-circuit to dc at steady state
Practice Problem 7.10 Find v(t) for t > 0. Assume the switch has
been open for a long time and is closed at t = 0. calculate at v(t) t = 0.5.
Ans: -5+15e-2tV, 0.5182V
/
/ 2 0.5
0.5
2
( ) ( ) [ (0) ( )]30 (15 30) (30 15 )1 (1) 30 15 20.9024 (4) 30 15 27.97
t
t t
v t v v v ee e V
At t s v e VAt t s v e V
= + = + = = = == = =
-
27
52
Example 7.11
In the figure, the switch has been closed for a long time and is opened at t = 0. Find i and v for all time.
Sol: By definition of the unit step function
Since the capacitor voltage cannot change
0, 030 ( )
30, 0For 0
10 , 110
tu t
tt
vv V i A
0, by using voltage division
/
(3/ 5) 0.6
0.6 0.6 0.6
20( ) (30) 2020 10
10 20 2010 2030 3
20 1 53 4 3
( ) ( ) [ (0) ( )]20 (10 20) (20 10 )
201 0.5 0.25( 0.6)( 10) (1 )
Th
Th
t
t t
t t t
v V
R
R C s
v t v v v ee e V
v dvi Cdt
e e e A
= =+= = =
= = == + = + =
= += + = +
-
28
54
Example 7.11
Practice Problem 7.11 The switch is closed at t = 0. Find
i(t) and v(t) for all time. Note that u(-t) = 1 for t < 0 and 0 for t > 0. Also u(-t) = 1- u(t).
Ans:
0.6
0.6
10 , 0(20 10 ) , 0
1 , 0(1 ) , 0
t
t
V tv
e V tA t
ie A t
55
7.6 Step Response of an RL Circuit
Consider the RL circuit, which may be replaced by the circuit applying unit step function
Our goal is to find the inductor current Decomposing the answer into transient and
steady-state responses, i.e. i = it + iss Since transient response is a decaying
exponential; that is
After long time, the inductor short-circuited
/ , , is a constantttLi Ae AR
= =
sss
ViR
=
-
29
56
Step Response of an RL Circuit
The complete response is Let I0 be the initial inductor current Since the current through the inductor cannot change
instantaneously, thus
At t = 0,
Therefore,
0(0 ) (0 )i i I+ = =
00 0 or s
V VI A A IR R
= + =
/0( ) ( )
ts sV Vi t I eR R
= +
/t sVi AeR
= +
57
Step Response of an RL Circuit
The response may be written as
where i(0) and i() are the initial and final values of i To find the step response of an RL circuit requires: The initial inductor current i(0) at t = 0 The finial inductor current i() The time constant
If the switching happens at time t = t0 instead of t = 0, then
/( ) ( ) [ (0) ( )] ti t i i i e = +
0( ) /0( ) ( ) [ ( ) ( )]
t ti t i i t i e = +
-
30
58
Step Response of an RL Circuit
If I0 = 0, then
The voltage across the inductor
/
/
0, 0( )
(1 ), 0
( ) (1 ) ( )
ts
ts
ti t V e t
RVor i t e u tR
=
/
/
( ) , , 0
( ) ( )
ts
ts
di L Lv t L V e tdt R R
v t V e u t
= = = >=
59
Example 7.12
Find i(t) in the circuit in the figure for t > 0. Assuming that the switch has been closed for a long time.
Sol:10(0 ) 52
(0) (0 ) (0 ) 50,
10( ) 22 3
2 3 51
135 15
Th
Th
i A
i i i AWhen t the switch is open
i A
R
L sR
+
= == = =
> = =+= + =
= = =
-
31
60
Example 7.12
/ 15
15
15 15
( ) ( ) [ (0) ( )] 2 (5 2)2 3 , 0:0
10 5
15 [10 15 ] [ (3)( 15) ] 103
t t
t
t t
Thusi t i i i e e
e A tCheckfor t KVL must be satisfied
dii Ldt
dii L e edt
= + = + = + >
>= +
+ = + + =
61
Practice Problem 7.12
The switch has been closed for a long time. It opens at t = 0. Find i(t) for t > 0.
Ans: (2 + e-10t) A, t > 0
-
32
62
Example 7.13
At t = 0, switch S1 in the figure is closed, and switch S2 is closed 4s later. Find i(t) for t > 0. Calculate i for t = 2s and t = 5s.
Sol: For t < 0 , For 0 t < 4
00i0i0i === + )()()(
( )4t0Ae14
e404ei0iiti
50105
RL
1064RA464
40i
t2
t2
tTh
Th
4
s2215
322
5RL
3226
624624R
72721130
6vi
11180v
6v
2v10
4v40
i4e144i4i
Th
8
Th
===
=+=+=
===
==+
==
||
A.)(
V
KCLUsing,)(findTo)()()(
-
33
64
Example 7.13
( )4)272.24(272.2
)()4()()()4(4667.1
/)4(
>+=+=
teeiiiti
t
t
2
1.4667( 4)
4
1.4667
0, 0( ) 4(1 ), 0 4
2.727 1.273 , 42
(2)
T
4(1 ) 3.935
(5) 2.727 1.273 3.02
o sum up,
t
t
ti t e t
e tAt ti e AAt ti e A
= + =
= ==
= + =
65
Practice Problem 7.13
Switch S1 is closed at t = 0, and switch S2 is closed at t = 2s. Calculate i(t) for all time. Find i(1) and i(3).
Ans:9
5( 2)
0, 0( ) 2(1 ), 0 2
3.6 1.6 , 2(1) 1.9997 , (3) 3.589
t
t
ti t e t
e ti A i A
= = =
-
34
66
7.7 First-Order Op Amp Circuits
An op amp circuit containing a storage element will exhibit first-order behavior
Differentiators and integrators are examples of first-order op amp circuits
For practical reasons, inductors are hardly used in op amp circuits; only RC type circuits are considered
The following three examples illustrate the location of the capacitor will influence the circuit behaviors; i.e. located in the input, output, or the feedback loop
67
Example 7.14 (source-free op amp circuit)
For the op amp circuit in the figure, find vo for t > 0, given that v(0)= 3V. Let Rf = 80 k, R1 = 20 kand C = 5 F.
Sol: Method 1:
1
1
11
/ 100 1
00
3 6 10 100
0Applying at 1 ,
0
( ) , , ( ) 3Applying at 2
0
80 10 5 10 ( 30 ) 12 , 0
t t
ff
t t
v dvKCL node CR dt
dv vv vdt CR
v t V e R C v t eKCL node
dv v dvC or v R Cdt R dt
v e e V t
=
= + == = =
= =
= = >
Q
-
35
68
Example 7.14
Method 2:
0
/0 0 0 0
10 10
Applying at 23 0 (0 ) 0, (0 ) 12
20,000 80000Applying to the input loop20,000(1) 0 20
2010.1. ,
( ) ( ) [ (0) ( )]
0 (12 0) 12 , 0
o
eq
eq
t
t t
KCL nodev v V
KVLv v kV
vThen R k
and R C Thus
v t v v v ee e V t
++
+ = =
= == =
= == + = + = >
69
Practice Problem 7.14
For the op amp circuit in the figure, find vo for t > 0, if v(0) = 4V. Assume that Rf = 50 k, R1 = 10 k and C = 10 F.
Ans: -4e-2t V, t > 0
-
36
70
Example 7.15
Determine v(t) and vo(t) in the circuit. Sol:
/
3 6
1
0 1
1 0
( ) ( ) [ (0) ( )] , 050 10 10 0.05
0, (0) 0200, 3 2
20 1050( ) (1 ) 3.5 2 720
So that ( ) 2 7 5
tv t v v v e tRC
For t v
For t v V
v v V
v v vv V
= + >
= = =< => = =+
= + = = =
= =
71
Example 7.15
Practice Problem 7.15 Find v(t) and vo(t) in the op amp circuit.
Ans: 40(1-e-10t) mV, 40(e-10t -1) mV
20 20
200 1
Substituting( ) 5 [0 ( 5)] 5( 1) , 0
We obtain ( ) ( ) ( ) 7 5 , 0
t t
t
v t e e V tv t v t v t e V t
= + = >
= = >
-
37
72
Example 7.16
Find the step response vo(t) for t > 0 in the op amp circuit. Let vi = 2u(t) V, R1 = 20 k,
R2 = R3 = 10 k, C = 2 F. Sol:
1
3 3
2 3 2 3 1
2 32 3 0
2 3
To find Thevenin voltage
To find Thevenin resistance
( 0)
fab i
fTh ab i
Th
RV v
R
RR RV V vR R R R R
R RR R R RR R
=
= = + +
= = =+
73
Example 7.16
3
2 3 1
2 3
2 3/
03 6
1000
10 50 2 ( ) 2.5 ( )20 20
5
( ) 2.5(1 ) ( )
where 5 10 2 10 0.01Thus the step response
( ) 2.5( 1) ( )
fTh i
Th
t
Th
t
RRV v u t u tR R RR RR k
R Rv t e u t
R C
v t e u t V
= = = += = +=
= = =
=
-
38
74
Practice Problem 7.16
Obtain the step response vo(t) in the op amp circuit. Let vi = 2u(t) V, R1 = 20 k, Rf = 40 k, R2 = R3 = 10 k, C = 2 F.
Ans: 6(1 - e-50t)u(t) V
75
7.9 Transient Analysis with PSpice
Transient response is the temporary response of the circuit thatsoon disappears
PSpice can be applied to obtain the transient response of a circuit with storage elements
If necessary, dc PSpice analysis is first carried out to determine the initial conditions
The initial conditions are used in the transient PSpice analysis to obtain the transient responses
It is recommended that during dc analysis, all capacitors are open-circuited while all inductors are short-circuited
-
39
76
Example 7.17
Use Pspice to find the response i(t) for t > 0 in the circuit.
77
Practice Problem 7.17
For the circuit, use Pspice to find v(t) for t > 0.
-
40
78
Example 7.18
In the circuit, use Pspice to find v(t) for t > 0.
Circuit for t > 0 Simplified circuit
( ) 10 18 tv t e V=
79
Example 7.18
Method 1 (dc transient analysis)
dc analysis
transient analysis
-
41
80
Example 7.18
Method 2 (simulate directly)
81
Practice problem 7.18
The switch was open for a long time but closed at t = 0. If i(0) = 10A, find i(t) for t > 0 by hand and also by Pspice.
Ans: i(t) = 6 + 4e-5t
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42
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7.9 Applications
Devices making use of RC or RL circuits include filtering in dc power supplies, smoothing circuits in digital communications, differentiators, integrators, delay circuits, and relay circuits
The short or long time constants of the RC or RL circuits are applied in the devices
In the following, delay circuits, photoflash unit, relay circuits and automobile ignition circuit are introduced
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7.9.1 Delay Circuits
An RC circuit can be used to provide various time delays
When the switch is closed, the capacitor voltage increase gradually toward 110V at a rate determined by the circuits time constant,(R1+R2)C
The lamp will not emit light until the voltage across it exceed 70V
When the voltage level is reached, the lamp fires and the capacitor discharge through it
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43
84
Delay Circuits
Due to the low resistance of the lamp when on, the capacitor voltage drops fast and the lamp turns off
Therefore, the lamp will turn on and off repeatedly Adjusting R2, we can introduce either short or long time delay The warning blinkers commonly found on road construction
sites are the application examples
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Example 7.19
Consider the circuit in the figure, and assume that R1 = 1.5 M, 0 < R2 < 2.5 k. (a) Calculate the extreme limits of the time constant of the circuit. (b) How long does it take for the first time after the switch is closed? Let R2 assume its largest value.
Sol: (a)
26 6
1 2
26 6
1 2
The samllest value for is 0
( ) (1.5 10 0) 0.1 10 0.15The largest value for is 2.5
( ) (1.5 2.5) 10 0.1 10 0.4
RR R C s
R MR R C s
= + = + =
= + = + =
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44
86
Example 7.19
(b)
0 0
0 0
/ /
/ /
/ /
0
(0) 0, ( ) 0
( ) ( ) [ (0) ( )] 110[1 ]0.4 , 70
770 110[1 ] 111
4 1111 4
Taking the natural logarithm of both sides11ln 0.4ln 2.75 04
c ct t
c c c c
c
t t
t t
v vv t v v v e ewhere s the lamp glows when v V
e e
or e e
t
= == + =
= == =
= =
= = = .4046s
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Example 7.19
Practice Problem 7.19 The RC circuit is designed to
operate an alarm which activateswhen the current through it exceeds
120A. If 0 R 6k, find the range of the time delay that the variable resistor can create.
Ans: Between 47.23ms and 124ms
0 00
0
0
(0) ( )A more general formula for finding is ln( ) ( )
The lamp will fire repeatedly every seconds if and only if ( ) ( )
v vt tv t v
tv t v
=
-
45
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7.9.2 Photoflash Unit
This application exploits the ability of the capacitor to oppose any abrupt change in voltage
The charging time is approximately tcharge = 5R1C(when the switch is in position 1)
The discharging time is approximately tdischarge = 5R2C (when the switch is in position 2)
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Photoflash Unit
This circuit provides a short-duration, high-current pulse
11
sVIR
=
22
sVIR
=
-
46
90
Example 7.20
An electron flashing has a current-limiting 6k resistor and 2000F electrolytic capacitor to 240V. If the lamp resistance is 12, find: (a) the peak charging current, (b) the time required for the capacitor to fully charge, (c) the peak discharging current, (d) the total energy stored in the capacitor, and (e) the average power dissipated by the lamp.
Sol: (a) The peak charging current is
(b)
(c) The peak discharging current is
(d) The energy stored is
1 31
240 406 10
sVI mAR
= = =3 6
charge 15 5 6 10 2000 10 60 1 minutet R C s= = = =
22
240 2012
sVI AR
= = =2 6 21 1 2000 10 240 57.6
2 2sW CV J= = =
91
Example 7.20
(e)
Practice problem 7.20 The flash unit of a camera has a 2mF capacitor charged to 80V.
(a) How much charge is on the capacitor?(b) What is the energy stored in the capacitor?(c) If the flash fires in 0.8ms, what is the average current through the flashtube?(d) How much power is delivered to the flashtube?(e) After a picture has been taken, the capacitor needs to be recharged by a power unit that supplies as maximum of 5mA. How much time does it take to charge the capacitor?
Ans: (a) 0.16C, (b) 6.4J, (c) 200A, (d) 8kW, (e) 32s
6discharge 2
discharge
5 5 12 2000 10 0.1257.6 4800.12
t R C sWp W
t
= = == = =
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7.9.3 Relay Circuits
A magnetically controlled switch is called a relay
The coil current gradually increases and produces a magnetic field
Eventually the magnetic field is sufficiently strong to pull the movable contact in the other circuit and close switch
At this time the relay is said to be pulled in The time interval td between the closure of
switch S1 and S2 is called the relay delay time
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Example 7.21
The coil of a certain relay is operated by a 12V battery. If the coil has a resistance of 150 and an inductance of 30mH and the current needed to pull in is 50mA, calculate the relay delay time.
Sol:
/
3
/
The current through the coil is given by( ) ( ) [ (0) ( )]
12where (0) 0, ( ) 80 ,150
30 10and 0.2150
Thus ( ) 80(1 )If ( ) 50 , then
t
t
d
i t i i i e
i i mA
L msR
i t e mAi t mA
= + = = =
= = ==
=
-
48
94
Example 7.21
Practice problem 7.21 A relay has a resistance of 200 and an inductance of 500mH. The relay
contacts close when the current through the coil reaches 350mA. What time elapses between the application of 110V to the coil and contact closure?
Ans: 2.529ms
/ /
/ /
550 80(1 ) 18
3 88 3
By taking the natural logarithm on both sides, we get8 8ln 0.2ln 0.19623 3
d d
d d
t t
t t
d
e e
or e e
t ms ms
= =
= =
= = =
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7.9.4 Automobile Ignition Circuit
The ability of inductors to oppose rapid change in current makes them useful for arc or spark generation
To form a spark requires thousands of volts between the air gap
How can such a high voltage be obtained from the car battery? Since the voltage across the inductor is v = Ldi/dt , we can make di/dt
large by creating a large change in current through the inductor
When the ignition switch is closed, the final current value isi = Vs / R
The time taken to charge is charge 5 5LtR
= =
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96
Automobile Ignition Circuit
At steady state, di/dt = 0, therefore, inductor voltage v = 0 When the switch suddenly opens, a large voltage is developed
across the inductor causing a spark The induced large voltage is due to the rapidly collapsing field The spark continues until the energy stored in the inductor is
dissipated in the spark discharge
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Example 7.22
A solenoid with resistance 4 and inductance 6mH is used in an automobile ignition circuit. If the battery supplies 12V, determine: the final current through the solenoid when the switch is closed, the energy stored in the coil, and the voltage across the air gap, assuming that the switch takes 1s to open.
Sol:
2 3 2
36
The final current through the coil is12 34
The energy stored in the coil is1 1 6 10 3 272 2
The voltage across the gap is36 10 18
1 10
sVI AR
W LI mJ
IV L kVt
= = =
= = =
= = =
-
50
98
Example 7.22
The spark coil of an automobile ignition system has a 20mH inductance and a 5 resistance. With a supply voltage of 12V, calculate: the time needed for the coil to fully charge, the energy stored in the coil, and the voltage developed at the spark gap if the switch opens in 2s.
Ans: 20ms, 57.6mJ, and 24kV
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Assignments
7.3, 7.7, 7.16, 7.19, 7.26, 7.39, 7.43, 7.49, 7.56, 7.66, 7.73