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Yesterday
• Ohm’s Law
V=IR
• Ohm’s law isn’t a true law but a good approximation for
typical electrical circuit materials
• Resistivity =1/ (Conductivity): Property of the material
• Resistance proportional to resistivity and length, inversely
proportional to area
L
R
A

Question 1
Two cylindrical resistors are made from
the same material, and they are equal in
length. The first resistor has diameter d,
and the second resistor has diameter 2d.
Compare the resistance of the two cylinders.
a) R1 > R2
b) R1 = R2
c) R1 < R2
Question 1
1. a
2. b
3. c
Question 1
Two cylindrical resistors are made from
the same material, and they are equal in
length. The first resistor has diameter d,
and the second resistor has diameter 2d.
Compare the resistance of the two cylinders.
a) R1 > R2
b) R1 = R2
c) R1 < R2
• Resistance is proportional to Length/Area
Question 2
Two cylindrical resistors are made from
the same material, and they are equal in
length. The first resistor has diameter d,
and the second resistor has diameter 2d.
If the same current flows through both resistors, compare the
average velocities of the electrons in the two resistors:
a) v1 > v2
b) v1 = v2
c) v1 < v2
Question 2
1. a
2. b
3. c
Question 2
Two cylindrical resistors are made from
the same material, and they are equal in
length. The first resistor has diameter d,
and the second resistor has diameter 2d.
If the same current flows through both resistors, compare the
average velocities of the electrons in the two resistors:
a) v1 > v2
b) v1 = v2
c) v1 < v2
Current  Area  Current Density
Current Density  average velocity of electrons
I is the same A1<A2
 v1>v2
Resistors in Series
•What is the same effective single
resistance to two resistances in series?
•Whenever devices are in SERIES, the
current is the same through both.
•By Ohm’s law, the Voltage difference
across resistance R1 is V  V  IR
a
•Across R2 is
b
1
I
a
R1
b
R2
c
a
Vb  Vc  IR2
•Total voltage difference
Va  Vc  I ( R1  R2 )
Reffective
• the effective single resistance is
Reffective  ( R1  R2 )
c
Another (intuitive) way…
R1
•Consider two cylindrical resistors
with lengths L1 and L2
L
V
R1   1
A
L2
R2  
A
L1
L2
R2
•Put them together, end to end to make a longer one...
L1  L2
Reffective  
 R1  R2
A
R  R1  R2
The World’s Simplest (and most useful) circuit:
Voltage Divider
R1
V ?
V
 V0 
V  IR2  
 R2
 R1  R2 
V0
R2
R2  R1
By varying R2 we can
controllably adjust
the output voltage!
R2  R1
R2  R1
V=0
V0
V=
2
V=V0
Question 3
Two resistors are connected in series to a
battery with emf E. The resistances are
such that R1 = 2R2. The currents through
the resistors are I1 and I2 and the potential
differences across the resistors V1 and V2.
Are:
a) I1>I2 and V2=E
b) I1=I2 and V2= E
c) I1=I2 and V2=1/3E
d) I1<I2 and V2=1/2E
e) I1<I2 and V2=1/3E
Resistors in Parallel
• Very generally, devices in parallel
have the same voltage drop
• Current through R1 is I1.
• Current through R2 is I2.
V  I1R1
a
V V V
 
R R1 R2

R2
I
I
a
V
1 1 1
 
R R1 R2
I2
R1
d
• But current is conserved

I1
V
V  I 2 R2
I  I1  I 2
I
R
d
I
Another (intuitive) way…
Consider two cylindrical resistors with
cross-sectional areas A1 and A2
L
R1  
A1
L
R2  
A2
V
A1
R1
Put them together, side by side … to make one
“fatter”one,
L
A1
A2
1
1
1





Reffective 
Reffective  L  L R1 R2
 A1  A2 

1 1 1
 
R R1 R2
A2
R2
Kirchhoff’s First Rule
“Loop Rule” or “Kirchhoff’s Voltage Law (KVL)”
"When any closed circuit loop is traversed, the algebraic
sum of the changes in potential must equal zero."
KVL:
V
n
0
loop
• This is just a restatement of what you already know:
that the potential difference is independent of path!
e1
I
e
 1
R1
 IR1
R2
 IR2
e2
e
 2 0
Rules of the Road
Our convention:
• Voltage gains enter with a + sign, and voltage drops enter
with a  sign.
• We choose a direction for the current and move around
the circuit in that direction.
• When a battery is traversed from the negative terminal to
the positive terminal, the voltage increases, and hence the
battery voltage enters KVL with a + sign.
• When moving across a resistor, the voltage drops, and
hence enters KVL with a  sign.
e1
e
 1
I
R1
R2
 IR1
 IR2
e2
e
 2
0
Current in a Loop
R1 e
1
Start at point a (could be
anywhere) and assume
current is in direction
shown (could be either)
b
V
n
f
a
I
I
d
c
R2
KVL:
R4
e2
e
R3
0
 IR1  IR2  e 2  IR3  IR4  e1  0

e1  e 2
I
R1  R2  R3  R4
loop
Question 3
• Consider the circuit shown.
– The switch is initially open and the current
flowing through the bottom resistor is I0.
– Just after the switch is closed, the current
flowing through the bottom resistor is I1.
– What is the relation between I0 and I1?
(a) I1 < I0
(b) I1 = I0
12V
R
a I
12V
12V
R
b
(c) I1 > I0
Question 3
1. a
2. b
3. c
Question 3
• Consider the circuit shown.
– The switch is initially open and the current
flowing through the bottom resistor is I0.
– Just after the switch is closed, the current
flowing through the bottom resistor is I1.
– What is the relation between I0 and I1?
(a) I1 < I0
(b) I1 = I0
12V
R
a I
12V
12V
R
b
(c) I1 > I0
• From symmetry the potential (Va-Vb) before the switch is
closed is Va-Vb = +12V.
• Therefore, when the switch is closed, potential stays the
same and NO additional current will flow!
• Therefore, the current before the switch is closed is
equal to the current after the switch is closed.
Question 3
• Consider the circuit shown.
– The switch is initially open and the current
flowing through the bottom resistor is I0.
– After the switch is closed, the current
12V
flowing through the bottom resistor is I1.
– What is the relation between I0 and I1?
(a) I1 < I0
•
•
(b) I1 = I0
Write a loop law for original loop:
12V +12V  I0R  I0R = 0
I0 = 12V/R
Write a loop law for the new loop:
12V  I1R = 0
I1 = 12V/R
12V
R
a
I
R
b
(c) I1 > I0
12V
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
How to use Kirchhoff’s Laws
A two loop example:
R1
I3
e1
e2
I2
I1
R2
R3
• Assume currents in each section of the circuit, identify all circuit
nodes and use KCL.
(1) I1 = I2 + I3
• Identify all independent loops and use KVL.
2 e1  I1R1  I2R2 = 0
(3) I2R2  e2  I3R3 = 0
4 e1  I1R1  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
e2
I2
I1
e1
e1 = 24 V
I3
R2
e 2 = 12 V
Then,
R3
R1= 5W
R2=3W
R3=4W
and
I1=2.809 A
I2= 3.319 A, I3= -0.511 A
Junction Demo
Junction:
e1
I1  I 2  I 3
R
Outside loop:
e1  I1R  I3 R  e3  0
I1
I2
e2
R
Top loop:
e1  I1R  e 2  I 2 R  0
2e1  e 2  e 3
I1 
3R
I2 
e1  e 3  2e 2
3R
I3
e3
R
e1  e 2  2e 3
I3 
3R
1
Summary
• Kirchhoff’s Laws
– KCL: Junction Rule (Charge is conserved)
– Review KVL (V is independent of path)
• Non-ideal Batteries & Power
• Discharging of capacitor through a Resistor:
Q (t )  Q0 e
t
RC
Reading Assignment: Chapter 26.6
Examples: 26.17,18 and 19
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
I
Four identical resistors are
connected to a battery as
shown in the figure.
R2
R1
R4
E
R3
5) How does the current through the battery change after the
switch is closed ?
a) Iafter > Ibefore
b) Iafter = Ibefore
c) Iafter < Ibefore
Before:
Rtot = 3R
Ibefore = 1/3 E/R
After:
R23 = 2R
R423 = 2/3 R
Rtot = 5/3 R
Iafter = 3/5 E/R