Transcript Lecture #12 03/01/05

Announcements
• Quiz II March 3rd
– Lectures 6-10
– Chapters 25,26, and 27 through Tuesday 02/22
– Slightly shorter: 10 MC and 3 SA
• There is a midsemester survey on webassign
• Most of you did not give an example on today’s
reading quiz.
Full Credit now, but not if it shows up
elsewhere!
• Office Hrs: Today
– 2-3pm
Batteries
•An ideal battery creates a voltage difference between the two
sides
E
P  I V
E2
P
R
–
+
R
If we can neglect the internal resistance in the battery, the current in
this simple circuit is just:
V 
I

R
R
Power Consumed by a
Resistor
V=E
+
E
–
R
I = E/R
P  I V
V=0
dU
d
dQ
P
  QV 
V  I V
dt
dt
dt
2
V 

2
P  RI 
R
Series Resistors
•Two or more resistors connected together can be treated as
one giant resistor
•For resistors in series, the current is the same through both of
them
R1
R2
V1 = R1 I
V2 = R2 I
V = V1+ V2 = R1 I + R2 I = (R1 + R2 )I
R = R1 + R2
Series Circuits
R1
R2
Resistors
V1 = R1 I
R = R1 + R2
V2 = R2 I
Current is the same through each resistor
Capacitors:
C1
C2
C3
1 1
1
1
 

C C1 C2 C3
Charges are the same on each capacitor
Real Batteries
•An real battery has an internal resistance
E
How does this affect the current?
  Ir  IR  0
I

Rr
–
r
+
R
real battery
Parallel Resistors
•Two or more resistors connected together can be treated as
one giant resistor
•For resistors in parallel, the voltage is the same across both of
them
V = R2 I2
R1
R2
V = R1 I1
1 1
1
 
R R1 R2
1 1 
V V
 V   
I  I1  I 2 

R1
R2
 R1 R2 
Parallel Circuits
Voltages are the same across each element
Resistors
1 1
1
 
R R1 R2
R1
Capacitors
C1
Same Voltages
C2
C3
C  C1  C2  C3
What is the combined resistance
of the mess at right?
A) 1.5 k
B) 12 k
C) 14 k
D) 21 k
9 k
3 k
3 k
6 k
3 k
2 k
R  R1  R2  3  3  6
1 1
1 1 1 1
 
  
R R1 R2 3 6 2
R  R1  R2  R3  3  9  2  14
3 k
Kirchoff’s Rules
•The total current flowing into a point must equal the total
current flowing out of a point [conservation of charge]
•The total voltage change around a loop must total zero
I2
I3
V1
V2
–
+
I1
I3=I2+I1
V3
V1 + V2 + V3 = 0
Using Kirchoff’s Rules
•Draw a (circuit) diagram and label everything known or
unknown!
•To every series of components, assign a direction to the
current I (don’t worry if you get it wrong, the result will be
correct just negative)
•You must be consistent however after you assign a
direction!
•Write down conservation of charge at each vertex
•Write down one equation for each loop
•In an emf source, going – to + gives a positive V, + to - is a
negative V
•Solve all equations
You might end up with many equations, but I trust that
you can solve simultaneous equations.
A Multiloop Circuit
I1 + I3 = I2
1.5 – 3I2 = 0
9 – 5I1 – 3I2 = 0
I2 = 1.5/3 = 0.5 A
I1 = (9 – 3I2)/5 = 1.5 A
I3 = I2 – I1 = 0.5 – 1.5 = – 1 A
I1
–
+
9V
5
I2
3
1.5 V
– +
I3
What
Whatisisthe
thevoltage
conservation
loop rule
of
you
current
get applied
law associated
to the upper
with
loop?
the junction on the right?
A)
A) 9I1++5II21 =+ I3I
32=0
B)
B) 9I+1 +5II13 –=3I
I22 = 0
C)
C) 9I–2 +5II13+=3I
I12 = 0
D)
D) 9I1–+5II21 +
– 3I
I3 2==00
A Multiloop Circuit
•There is one more loop in the
problem.
9 – 5I1 -1.5 = 0
I1 = (9 – 1.5)/5 = 1.5 A
I1
–
+
9V
5
I2
3
1.5 V
– +
I3
•We only had one resistor and so only had to consider
one current. This can simplify problems!
Odd Circuit
What is the current through the
resistor?
A) 3.6 A
B) 1.8A
C) 90 A
D) 0 A
–
+
9V
5
9V
– +