Transcript Short Version : 25. Electric Circuits

Short Version :
25. Electric Circuits
Electric Circuit = collection of electrical components connected by conductors.
Examples:
Man-made circuits: flashlight, …, computers.
Circuits in nature: nervous systems, …, atmospheric circuit (lightning).
25.1. Circuits, Symbols, & Electromotive Force
Common circuit symbols
All wires ~ perfect conductors  V = const on wire
Electromotive force (emf) = device that maintains fixed V across its terminals.
E.g.,
batteries (chemical),
generators (mechanical),
photovoltaic cells (light),
cell membranes (ions).
m~q
Collisions ~ resistance
g~E
Ideal emf :
no internal energy loss.
Lifting ~ emf
Energy gained by charge transversing battery = q E
( To be dissipated as heat in external R. )
Ohm’s law: I  E
R
25.2. Series & Parallel Resistors
Series resistors :
I = same in every component
Same q must go
every element.
E  V1  V2  I R1  I R2  I Rs

Rs  R1  R2
n
For n resistors in series:
I
n=2:
V1 
R1
R1  R 2
E
E
Rs
Rs   R j
Vj  I Rj 
V2 
R2
R1  R 2
E
j 1
Rj
Rs
E
Voltage divider
Real Batteries
Model of real battery = ideal emf E in series with internal resistance Rint .
I means V drop I Rint
 Vterminal < E
E  I Rint  I RL
I
E
Rint  R L
VRL 
RL
R int  R L

E
E
RL
Example 25.2. Starting a Car
Your car has a 12-V battery with internal resistance 0.020 .
When the starter motor is cranking, it draws 125 A.
What’s the voltage across the battery terminals while starting?
Battery terminals
E  I R int  I RL
RL 
E
12 V
 Rint 
 0.020 
I
125 A
 0.096   0.020 
 0.076 
Voltage across battery terminals = E  Vint  12 V  125A  0.020   9.5V
Typical value for a good battery is 9 – 11 V.
Parallel Resistors
Parallel resistors :
V = same in every component
I  I1  I 2

E
E
E



Rp
R1 R 2
1
1
1


R p R1 R 2
For n resistors in parallel :
Rp 
R1 R 2
R1  R 2
n
1
1

Rp j  1 R j
Analyzing Circuits
Tactics:
• Replace each series & parallel part by their single component equivalence.
• Repeat.
Example 25.3. Series & Parallel Components
Find the current through the 2- resistor in the circuit.
Equivalent of parallel 2.0- & 4.0- resistors:
1
1
1
3



R 2.0  4.0 
4.0 

R  1.33 
Equivalent of series 1.0-, 1.33-  & 3.0-
resistors:
RT  1.0   1.33   3.0   5.33 
Total current is
I5.33 
12 V
E

RT 5.33 
 2.25 A
Voltage across of parallel 2.0- & 4.0- resistors: V1.33   2.25 A1.33   2.99 V
Current through the 2- resistor:
I 2 
2.99 V
2.0 
 1.5A
25.3. Kirchhoff’s Laws & Multiloop Circuits
Kirchhoff’s loop law:
 V = 0 around any closed loop.
( energy is conserved )
Kirchhoff’s node law:
This circuit can’t be analyzed using
series and parallel combinations.
I=0
at any node.
( charge is conserved )
Multiloop Circuits
Problem Solving Strategy:
INTERPRET
■
Identify circuit loops and nodes.
■
Label the currents at each node, assigning a direction to each.
DEVELOP
■
Apply Kirchhoff ‘s node law to all but one nodes. ( Iin > 0, Iout < 0 )
■
Apply Kirchhoff ‘s loop law all independent loops:
Batteries: V > 0 going from  to + terminal inside the battery.
Resistors: V =  I R going along +I.
 Some of the equations may be redundant.
Example 25.4. Multiloop Circuit
Find the current in R3 in the figure below.
Node A:
 I1  I 2  I 3  0
Loop 1:
E1  I1R1  I 3R3  0
6  2 I1  I 3  0
Loop 2:
E2  I 2 R2  I3R3  0
9  4I2  I3  0

1
I1   I 3  3
2
I2 
9
1 1 


1
I

3


 3
4
2 4 

I3 
4 21

 3A
7 4
1
9
I3 
4
4
Application: Cell Membrane
Hodgkin-Huxley (1952) circuit model of cell membrane (Nobel prize, 1963):
Resistance of cell membranes
Membrane
potential
Electrochemical effects
Time dependent effects
25.4. Electrical Measurements
A voltmeter measures potential difference between its two terminals.
Ideal voltmeter: no current drawn from circuit  Rm = 
Example 25.5. Two Voltmeters
You want to measure the voltage across the 40- resistor.
What readings would an ideal voltmeter give?
What readings would a voltmeter with a resistance of 1000  give?
(a)
(b)


40 
V40   
12 V   4V

 40   80  
Rparallel 
 40 1000 
40   1000 
 38.5 


38.5 
V40  
 12 V   3.95V
 38.5   80  
Ammeters
An ammeter measures the current flowing through itself.
Ideal voltmeter: no voltage drop across it  Rm = 0
Ohmmeters & Multimeters
An ohmmeter measures the resistance of a component.
( Done by an ammeter in series with a known voltage. )
Multimeter: combined volt-, am-, ohm- meter.
25.5. Capacitors in Circuits
Voltage across a capacitor cannot change instantaneously.
The RC Circuit: Charging
C initially uncharged  VC = 0
Switch closes at t = 0.
VR (t = 0) = E

I (t = 0) = E / R
C charging:
VR  but rate 
I  but rate 
VC  but rate 
VC   VR   I 
Charging stops when I = 0.
E I R

dI
dt

I
RC


Q
0
C
dI
I
R 0
dt
C
I
dQ
dt
t dt
dI


I0 I 0 RC
I
VC ~ 2/3 E
ln
I
t

I0
RC
I  I0 e

t
RC
VC  E  VR
E  RCt
 e
R
t



 E  1  e RC 


Time constant = RC
I ~ 1/3 E/R
The RC Circuit: Discharging
C initially charged to VC = V0
Switch closes at t = 0.
VR = VC = V

Q
I R0
C
I 
dI
dt

I
RC
I  I0 e

t
RC
V  V0 e
V0  RCt
 e
R

t
RC
dQ
dt
I 0 = V0 / R
C discharging:
VC   VR   I 
Disharging stops when I = V = 0.
Example 25.6. Camera Flash
A camera flash gets its energy from a 150-F capacitor & requires 170 V to fire.
If the capacitor is charged by a 200-V source through an 18-k resistor,
how long must the photographer wait between flashes?
Assume the capacitor is fully charged at each flash.
 V 
t   RC ln  1  C 
E 

 170 V 
  18  103  150  106 F  ln 1 

200
V


 5.1 s
RC Circuits: Long- & Short- Term Behavior
For t << RC: VC  const,

C replaced by short circuit if uncharged.

C replaced by battery if charged.
For t >> RC: IC  0,

C replaced by open circuit.
Example 25.7. Long & Short Times
The capacitor in figure is initially uncharged.
Find the current through R1
(a) the instant the switch is closed and
(b) a long time after the switch is closed.
(a)
(b)
I1 
I1 
E
R1
E
R1  R2