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Today…
• Finish Kirchhoff’s Laws
– KCL: Junction Rule (Charge is conserved)
– Review KVL (V is independent of path)
• Non-ideal Batteries & Power
– What limits the current in a real battery
– How to calculate power
– Why we use high-voltage power lines
• Resistor-Capacitor Circuits, qualitative
Resistors in Parallel
• What to do?
V  IR
• Very generally, devices in parallel
have the same voltage drop
• But current through R1 is not I !
Call it I1. Similarly, R2 I2.
KVL 
V  I1R1  0
I
a
I1
V
I2
R1
I
d
V  I 2 R2  0
• How is I related to I 1 & I 2 ?
R2
I
a
Current is conserved!
V
I  I1  I 2

V V
V


R R1 R2

1 1 1
 
R R1 R2
R
d
I
Another (intuitive) way…
Consider two cylindrical resistors with
cross-sectional areas A1 and A2
L
R1  
A1
A1
V
A2
R1
L
R2  
A2
R2
Put them together, side by side … to make one “fatter”one,
Reffective 
L
 A1  A2 


1
Reffective
1 1 1
 
R R1 R2

A1
A
1
1
 2 

L L R1 R2
1
Lecture 10, ACT 1
• Consider the circuit shown:
1A
50W
a
– What is the relation between Va -Vd
and Va -Vc ?
b
12V
(a) (Va -Vd) < (Va -Vc)
(b) (Va -Vd) = (Va -Vc)
(c) (Va -Vd) > (Va -Vc)
1B
1B
(b) I1 = I2
20W
80W
d
– What is the relation between I1 and I2?
(a) I1 < I2
I2
I1
(c) I1 > I2
c
Lecture 10, ACT 1
• Consider the circuit shown:
1A
50W
a
– What is the relation between Va -Vd
and Va -Vc ?
(a) (Va -Vd) < (Va -Vc)
(b) (Va -Vd) = (Va -Vc)
(c) (Va -Vd) > (Va -Vc)
b
I2
I1
12V
20W
80W
d
c
• Do you remember that thing about potential being independent of path?
Well, that’s what’s going on here !!!
(Va -Vd) = (Va -Vc)
Point d and c are the same, electrically
Lecture 10, ACT 1
• Consider the circuit shown:
1A
50W
a
– What is the relation between Va -Vd
and Va -Vc ?
b
12V
(a) (Va -Vd) < (Va -Vc)
(b) (Va -Vd) = (Va -Vc)
(c) (Va -Vd) > (Va -Vc)
1B
1B
20W
• Note that: Vb -Vd
• Therefore,
(b) I1 = I2
(c) I1 > I2
= Vb -Vc
I1 (20W)  I 2 (80W)
80W
d
– What is the relation between I1 and I2?
(a) I1 < I2
I2
I1
I1  4I 2
c
Kirchhoff’s Second Rule
“Junction Rule” or “Kirchhoff’s Current Law (KCL)”
• In deriving the formula for the equivalent resistance of
2 resistors in parallel, we applied Kirchhoff's Second
Rule (the junction rule).
"At any junction point in a circuit where the current
can divide (also called a node), the sum of the
currents into the node must equal the sum of the
currents out of the node."
I in   I out
• This is just a statement of the conservation of charge at any
given node.
• The currents entering and leaving circuit nodes are known as
“branch currents”.
• Each distinct branch must have a current, Ii assigned to it
Old Preflight
Two identical light bulbs are
represented by the resistors
R2 and R3 (R2 = R3 ). The
switch S is initially open.
2) If switch S is closed, what happens to the brightness of the bulb R2?
a) It increases
b) It decreases
c) It doesn’t change
3) What happens to the current I, after the switch is closed ?
a) Iafter = 1/2 Ibefore
b) Iafter = Ibefore
c) Iafter = 2 Ibefore
How to use Kirchhoff’s Laws
A two loop example:
R1
I3
e1
e2
I2
I1
R2
R3
• Analyze the circuit and identify all circuit nodes
and use KCL.
(1) I1 = I2 + I3
• Identify all independent loops and use KVL.
(2) e1  I1R1  I2R2 = 0
(3) e1  I1R1  e2  I3R3 = 0
(4) I2R2  e2  I3R3 = 0
How to use Kirchoff’s Laws
R1
I3
e2
I2
I1
e1
R2
R3
• Solve the equations for I1, I2, and I3:
First find I2 and I3 in terms of I1 :
I 2  (e1  I1R1 ) / R2
From eqn. (2)
I3  (e1  e 2  I1R1 ) / R3
Now solve for I1 using eqn. (1):
I1 
e1
R2

e1  e 2
R3
R1 R1
 I1 (  )
R2 R3
e1

R2

From eqn. (3)
e1  e 2
R3
I1 
R R
1 1  1
R2 R3
Let’s plug in some numbers
R1
I1
e1
e1 = 24 V
Then
I3
e2
I2
R2
e 2 = 12 V
R3
R1= 5W
I1=2.809 A
The sign means that
the direction of I3 is
opposite to what’s
shown in the circuit
R2=3W
R3=4W
and
I2= 3.319 A
and
I3= 0.511 A
See Appendix for a more complicated example, with three loops.
Summary of Simple Circuits
• Resistors
in series: Reffective  R1  R2  R3  ...
Current thru is same;
• Resistors
1
in parallel: R
effective
Voltage drop across is IRi

1 1 1
   ...
R1 R2 R3
Voltage drop across is same;
Current thru is V/Ri
Kirchhoff’s laws: (for further discussion see online “tutorial essay”)
V
n
0
loop
I in   I out
Batteries
(“Nonideal” = cannot output arbitrary current)
I
I
• Parameterized with
R
"internal resistance"
V  e  Ir
r
V
e
e  Ir  IR  0
e
I
Rr

R
V e
Rr
Internal Resistance Demo
As # bulbs increases, what happens to “R”??
I
r
V
R
e
How big is “r”?
Power
Batteries & Resistors
Energy expended
chemical
to electrical
to heat
Rate is:
energy
 power  sJ 
time
What’s happening?
Assert:
Charges per time
P  VI
Potential difference per charge
or you can write it as
For Resistors:
Units okay?
P  IR I  I R
2
P  V V R   V 2 R
Joule  Coulomb J  Watt
Coulomb second s
Power Transmission
• Why do we use “high tension” lines to transport power?
–Transmission of power is typically at very high voltages (e.g., ~500 kV)
• But why?
–Calculate ohmic losses in the transmission lines
–Define efficiency of transmission:
Pout IVin  I 2 R
Pin R
IR  Vin 


 1
   1 2
Pin
IVin
Vin  Vin 
V in
Keep R
small
Make
Vin big
– Note: for fixed input power and line resistance, the inefficiency  1/V2
Example: Quebec to Montreal
1000 km  R= 220W
suppose Pin = 500 MW
With Vin=735kV, e = 80%.
The efficiency goes to zero quickly if
Vin were lowered!
Why do we use AC (60 Hz)? Easy to generate high voltage
(water/steam → turbine in magnetic field → induced EMF) [Lecture 16]
We can use transformers [Lecture 18] to raise the voltage for
transmission and lower the voltage for use
Resistor-capacitor circuits
I
Let’s try to add a Capacitor to
our simple circuit
I
R
Recall voltage “drop” on C?
V
e
Q
C
C
Q
Write KVL:
e  IR 
0
C
What’s wrong here?
Consider that I 
dQ
dt
and substitute. Now eqn. has only “Q”:
KVL gives Differential Equation !
dQ Q
eR
 0
dt C
We will solve this next time. For now, look at qualitative behavior…
Capacitors Circuits, Qualitative
Basic principle: Capacitor resists rapid change in Q 
resists rapid changes in V
• Charging (it takes time to put the final charge on)
– Initially, the capacitor behaves like a wire (DV = 0, since Q = 0).
– As current starts to flow, charge builds up on the capacitor
 it then becomes more difficult to add more charge
 the current slows down
– After a long time, the capacitor behaves like an open switch.
• Discharging
– Initially, the capacitor behaves like a battery.
– After a long time, the capacitor behaves like a wire.
2
Lecture 10, ACT 2
2A
• At t=0 the switch is thrown
from position b to position a in
the circuit shown: The
capacitor is initially
uncharged.
a
2B
(b) I0+ = e /2R
b
(c) I0+ = 2e /R
long time?
(b) I = e /2R
C
e
– What is the value of the current I after a very
(a) I = 0
I
R
– What is the value of the current I0+
just after the switch is thrown?
(a) I0+ = 0
I
(c) I > 2e /R
R
Lecture 10, ACT 2
2A
• At t=0 the switch is thrown
from position b to position a in
the circuit shown: The
capacitor is initially
uncharged.
a
(b) I0+ = e /2R
I
R
b
C
e
– What is the value of the current I0+
just after the switch is thrown?
(a) I0+ = 0
I
R
(c) I0+ = 2e /R
•
Just after the switch is thrown, the capacitor still has no
charge, therefore the voltage drop across the capacitor = 0!
•
Applying KVL to the loop at t=0+, e IR  0  IR = 0  I = e /2R
Lecture 10, ACT 2 a
2A
• At t=0 the switch is thrown from
position b to position a in the circuit
shown: The capacitor is initially
uncharged.
– What is the value of the current I0+
just after the switch is thrown?
I
I
R
b
C
e
R
(a) I0+ = 0
2B
(b) I0+ = e /2R
(c) I0+ = 2e /R
– What is the value of the current I after a very
long time?
(a) I = 0
(b) I = e /2R
(c) I > 2e /R
• The key here is to realize that as the current continues to flow, the
charge on the capacitor continues to grow.
• As the charge on the capacitor continues to grow, the voltage across
the capacitor will increase.
• The voltage across the capacitor is limited to e ; the current goes to 0.
Summary
• Kirchhoff’s Laws
– KCL: Junction Rule (Charge is conserved)
– Review KVL (V is independent of path)
• Non-ideal Batteries & Power
– Effective “internal resistance” limits current
– Power generated ( I V ) = Power dissipated ( I R  V
2
2
R)
– Power transmission most efficient at low current  high voltage
• Resistor-Capacitor Circuits
– Capacitors resist rapid changes in Q  resist changes in V
Appendix: A three-loop KVL example
• Identify all circuit nodes these are where KCL eqn’s
are found
I3
I1
I2
– determine which KCL
equations are algebraically
independent (not all are in
this circuit!)
–
–
–
–
I1=I2+I3
I4=I2+I3
I4=I5+I6
I1=I5+I6
I1=I4
I1=I2+I3
I4+I5+I6
• Analyze circuit and identify
all independent loops where
S DVi = 0 < KVL
I6
I4
I5
A three-loop KVL example
• Here are the node equations from
applying Kirchoff’s current law:
I1=I4
I1=I2+I3
I4+I5+I6
• Now, for Kirchoff’s voltage law:
(first, name the resistors)
I6R6+I5R5=0
I2R2b+I2R2a- I3R3 =0
VB-I1R1-I2R2a- I2R2b-I4R4-I5R5 = 0
Six equations, six unknowns….
I3
R1
I1
I2
R3
R2b
I6
R6
R5
• There are simpler ways of
analyzing this circuit, but this
illustrates Kirchoff’s laws
R2a
I5
R4
I4