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

7

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!

10

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.

17

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