14 fe review - rc and rl analysisrljones/fe_pe/14_fe review - rc and rl analysis.pdf · fe review...
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FE Review
RC and RL Analysis
1
Series CapacitorsTreat like parallel resistors
C 1 C 3C 21
200 f 50 f 10 f
32
6 0 v200 f 50 f 10 f
C T
2T 1 3
1 1 1C C C
1C
2T 1 3
T
1 1 1 1 125kC 200 f 50 f 10 f
2TC 8 f
Parallel CapacitorsTreat like series resistors
4 8 v 800 fC 1
60 fC 2 C 3
1200 f4 8 v 800 f 60 f 1200 fC T
1T 2 3C C CC
800 f 60 f 1200 2.0 ff 6m
3
Capacitor Current Divider(Like a resistor voltage divider)
I 2
5 A 8fC 1
6 fC 2
I 1
2
1 5A 8 fICI 2 857A
1
2
1
1 2
2
C C 8 f 6 f5A 6 fIC
C C 8 f 6 f
I 2.857A
I 2.143A 2
1 2C C 8 f 6 fThe two currents together add up to 5 amps (Naturally) (taking
MUST
4
add up to 5 amps (Naturally) (takingroundoff error into account)
Capacitor voltage divider(Like a resistor current divider)
5v 8 f5vC 40v 12
1 2
5v 8 f5vC 40vC C 8
V 2.856
7f 4
vf 1
5
InductorsInductors
Treat inductors just like resistors.This includes the inductive voltage
d t di id tiand current divider equations.
6
First Order FundamentalsFirst Order FundamentalsFinal ValueFinal Value
(steady state value)
t
v( ) v v( ) v et 0
Initial Value timeconstant
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Initial and Final ValuesInitial and Final Values
To start with let’s consider a switch InTo start with, lets consider a switch. Inanalyzing the switch, we will consider twoextremes: extremes: ‚ The instant a switch is closed (the
Initi l V lu ) ndInitial Value), and‚ the steady state value (a long time
has passed since the switch closed)has passed since the switch closed)(The Final Value).
8
When you have a circuit inwhich you are working out theinitial values,,! An uncharged capacitor
acts as a short circuitacts as a short circuit.(It opposes a change in
lt )voltage)! A un-fluxed inductor acts
as an open circuit.(It opposes a change in
9
( pp gcurrent).
Extending this idea to the device with anExtending this idea to the device with anInitial charge, we get:
C oV ( ) V a voltage source!0
oLi ( ) I a current sourc0 e!
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Final ConditionsFinal Conditions openCapacitor current 0ti ( ) 0
c
L
openshort
th t it th
Capacitor current 0tInduc
l ld b
i ( ) 0v ( ) 0
tor voltage 0t
c L
another way to write these values would b i ( ) 0
e: an v ( ) 0d
C Lo p e n s h o r tC Lp s h o r t
11
RC circuit – Charge PhaseRC circuit – Charge Phase
+Rt = 0
RC+- Cv s RC
12
Cy V (t)
C
c c c ct
y V (t)
V ( ) V V ( ) V e
d i d( ) ( )
t 0
S
C
Ct
determinedy( ) V ( )earliery V ( )
0 0V0v
S Sc
t
t
RC
RC
V ( ) e
V ( )
Vt
t 1
V
e
0
V
v
C SRCV ( )t 1 eV
13
( 9 8 2 )V
0.80.9
1
( . 8 6 5 )
( . 9 5 )( . 9 8 2 )
( . 9 9 3 )
V S
R C
0.50.60.7
( . 6 3 2 )
R Cn e t w o r kC H A R G EC
(t)
0.20.30.4
C H A R G Ep h a s e
VC
0.1
0 1 2 3 4 5
14( t a u )
RC Circuit – Discharge PhaseRC Circuit – Discharge Phase
c c c ct
Discharge phase
t 0
- Voltage
V ( ) V V ( ) V e
C CSV 0vNote that it is assumed here that the capacitorh d ti t l t l h t
0V ( ) V ( )
V
c C
s
S S
t tRC RC
had time to completely charge to
V 0v 0v( ) e V ( )t t e
V.
V V
15
0 .8
1
Vc(
t) VS
VRC voltagewaveform
0 .4
0 . 6
RC = 1
VS
VS
VS
0 .2
S
VS
00 2 4 6 8 1 0
t, s
1
0
VS
R RC current
0
0 . 20 . 40 . 60 . 8
(t) t, s
R
RC = 1
C cu e twaveform
- 0 . 8- 0 . 6- 0 . 4- 0 . 20ic
(
0 2 4 6 8 1 0
16- 1
0 . 8
ExampleExample
At time t = 0 the switch goes to position #1.The initial charge on the capacitor is 0 volts.g pShow the charge graph of the capacitor voltage.
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Step 1: Remove the load (the capacitor) and theDischarge Resistor and determine the Thevenin CktDischarge Resistor and determine the Thevenin Ckt.
Th 10k || 6k 4kR Th 10k || 6k 4k3.75k
R
7
4
5
k
7 k
18
7. 57 k
Th
5v 6kV
Th 1
V
1 875v0k 6k
19
1.875v
Place the Cap onto the Thevenin circuit and then calculate the charge circuit and then calculate the charge
eqn.
c c c ct
V ( ) V V ( ) V et 0
c ct
V ( ) 0v V 1.875v
RC 7.75k 8 f 62m
0
s
c
t
t
62ms
62ms
V ( ) 1.875v 0v 1.875v e
1.875 1.875e
t
20t62ms1.8 5 1 e7
Now place the cap with the discharge resistor and recalculate discharge resistor and recalculate
the eqn.
c c c ct
V ( ) V V ( ) V et 0
c cV ( ) 1.0 875v V 0v
RC 5k 8 f 40ms
c c
ct
ttV ( ) 0v 1.875v 0v e
21c
t40msV ( ) 1.875vet