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Resistors in series and in parallel (review)
I
R1
I1
R2
R1
I
V1
I  I1  I 2
I
V2
V
I2
V  V1  V2
Example
R1
R2
1
1
1


Req 15 30
R2
r  2
R1  15
E
R2  30
I ?
1
1 1
 
Req R1 R2
R1
  12V
V  V1  V2
I  I1  I 2
Req  R1  R2
Req  10
r
Req  15  30

R2
12V
I

 0.26 A
R1  R1  r 47
E
r

12V
I

 1.0 A
Req  r 12
10. Kirchhoff’s rules
1) Kirchhoffe’s first rule
(Conservation of charges and junction or current rule)
Example
I1
 I  I
I  I
in
I2
I1=I2+I3
in
I3
I1  5 A
5 A  2 A  I3
I2  2A
I3  3A
I3  ?
I 0
I in  0
I out  0
out
out
0
2) Kirchhoffe’s second rule
(Loop or voltage rule)
   I
n
m
E1 E2
Rm
R1
R2
E3
R3
E4<0
I
V
3) Using Kirchhoffe’s rules
 I  I
out
   I
m
in
n
Rm
For N junctions write N-1 equations
Write equations only for independent loops
Example 1: Determine the currents through the elements of this circuit.
I1
ε2
R1
I3
I2
ε1
• We have 3 different currents
• We have 2 junctions (N=2)
• We have 2 independent loops
R2
R3
We need 3 equations
We can use N -1=1 equation
We can use 2 equations
Example 1 (continuation)
Junction equation:
(Both junction give
the same equation)
Loop equations:
An extra loop equation:
If we add two equations above, we obtain the equation
for the third (big, combined) loop. We do not need this
equation – it is dependent from two previous
Let us use numbers:


1
 24V
2
 12V
R1  5
R2  3
R3  4
I1  I 2  I 3  0
 I R I R
  I R  I R
   I R  I R
1
1
2
3
1
2
1
2
3
1
2
1
2
2
3
We have three equations for three unknowns,
we can solve this system of equations.
I 1  2.8 A
I 2  3.3 A
I 3  0.5 A
I3 flows opposite to
our assumption
3
Example 2:

ε
1
 9.0V
2
 6.0V
E1
I1
R1
 I  I
in
R2
   I
I2
R1  30
R2  15
I ?
 
1

2
2
 I1 R1
 I 2 R2
I1  I 2  I 3
n
E2
I1
ε


I2


1
R1
2
Rm
9.0V  6.0V

0.5 A
30
6.0V

 0.4 A
R2 15
2
9 V
I 3  I 1  I 2  0.9 A
B. Decreases
m
I3
Example 3: In the circuit below, the switch is
initially open. When the switch is closed,
the current through the bottom resistor:
A. Increases
out
C. Stays the same
R
9 V
9 V
R
11. Microscopic picture
L
+
vd
A
For electrons in copper :
drift speed ~ 10-5 - 10-6 m/s
random-motion speed ~ 106 m/s
L  vd t
Q  nALe  neAvd t
Q
I
t
I  neAvd
n – number of electric carriers per unit volume
A – area of cross-section
AL - volume
Example: A copper wire 2.0 mm in diameter, caries 1.0 A. What is the
drift speed of electrons? Assume one electron per atom is free to move.
I  1. 0 A
r  2.0mm / 2  1.0mm
N A  6.02 10 23 mol 1
mmol  63.5 10 3 kg / mol
  8.9 103 kg / m 3
vd  ?
I
vd 
neA
I  neAvd
NA
NA
n

 8.24 10 28 m 3
Vmol mmol / 
A  r 2
I
1.0 A
vd 

neA 8.24 1028 m 3 1.6 1019 C   10 3 m 2

vd  2.4 10 5 m / s
