Resistors in Series
•When two or more resistors are connected end-toend, they are said to be in series.
•For a series combination of resistors, the currents
are the same in all the resistors because the
amount of charge that passes through one resistor
must also pass through the other resistors in the
same time interval.
•The potential difference will divide among the
resistors such that the sum of the potential
differences across the resistors is equal to the total
potential difference across the combination.
Section 28.2
Resistors in Series, cont
•Currents are the same
– I = I1 = I2
•Potentials add
– ΔV = V1 + V2 = IR1 + IR2
= I (R1+R2)
– Consequence of
Conservation of Energy
•The equivalent resistance
has the same effect on the
circuit as the original
combination of resistors.
Section 28.2
Equivalent Resistance – Series
•Req = R1 + R2 + R3 + …
•The equivalent resistance of a series
combination of resistors is the algebraic
sum of the individual resistances and is
always greater than any individual
resistance.
•If one device in the series circuit creates
an open circuit, all devices are inoperative.
Section 28.2
Equivalent Resistance
– Series – An Example
•Are all 3 representations equivalent?
A. yes
B. no
Two resistors are replaced with their equivalent
resistance.
Section 28.2
Resistors in Parallel
•The potential difference across each resistor
is the same because each is connected
directly across the battery terminals.
ΔV = ΔV1 = ΔV2
•A junction is a point where the current can
split.
•The current, I, that enters junction must be
equal to the total current leaving that junction.
– I = I 1 + I 2 = (ΔV1 / R1) + (ΔV2 / R2)
–The currents are generally not the same.
Section 28.2
–Consequence of conservation
of electric charge
Equivalent Resistance –
Parallel, Examples
•Are all three diagrams equivalent?
•Equivalent resistance replaces the two
original resistances.
Section 28.2
– Parallel Resistors
•Equivalent Resistance
1
1
1
1




Req R1 R2 R3
•The inverse of the
equivalent resistance of two
or more resistors connected
in parallel is the algebraic
sum of the inverses of the
individual resistance.
– The equivalent is always less
than the smallest resistor in
the group.
Section 28.2
Resistors in Parallel: Final Observations
•In parallel, each device operates independently of
the others so that if one is switched off, the others
remain on.
•In parallel, all of the devices operate on the same
voltage.
•The current takes all the paths.
– The lower resistance will have higher currents.
– Even very high resistances will have some currents.
•Household circuits are wired so that electrical
devices are connected in parallel.
Section 28.2
Effective resistance: two resistors in series
R1
R2

R1 and R2 experience the same
current but different voltages.
Largest R has largest V.
V V1  V2 I R1  I R2
Req  

 R1  R2
I
I
I
Req is larger than
either R1 or R2.
Effective resistance: two resistors in parallel
R1

R2
R1 and R2 experience the same
voltage but different currents.
Smallest R has largest I.
Req 
1
V
V
V
1 


   
I I1  I 2 V / R1  V / R2  R1 R2 
1
Req is smaller
than smallest
of R1 and R2.
Some Circuit Notes
•A local change in one part of a circuit may
result in a global change throughout the circuit.
–For example, changing one resistor will affect the
currents and voltages in all the other resistors and
the terminal voltage of the battery.
•In a series circuit, there is one path for the
current to take.
•In a parallel circuit, there are multiple paths
for the current to take.
Section 28.2
The switch is initially open. When the switch is closed,
the current measured by the ammeter will:
A. increase B. decrease C. stay the same D. fall to zero.

iClicker Quiz
(1) I have completed at least 50% of the reading and
study-guide assignments associated with the lecture,
as indicated on the course schedule.
A. True
B. False
Hint: This is a good time to read the chapter summaries about
circuits. BTW Current is NEVER used up.
Today: Review of series & parallel use video demos and MU.
How to add up resistors in series and parallel: light
bulb problems.
A. No new Circuit elements. B. Terms: Branch, junction
Loop C. Kirchhoff’s Laws. Pay attention to signs of
quantities and directions of current flow.
Combinations of Resistors
•The 8.0-W and 4.0-W
resistors are in series and can
be replaced with their
equivalent, 12.0 W
•The 6.0-W and 3.0-W
resistors are in parallel and
can be replaced with their
equivalent, 2.0 W
•These equivalent resistances
are in series and can be
replaced with their equivalent
resistance, •14.0
Section 28.2
W
Which part involves doing an integral?
A. a
B. b
C. both a and b.
D. neither. E. Squirrel
Resistance of an object with arbitrary shape
Inside
to out
End-to-end:
dz

 length 
dR  d  

dz

2
2
A(z )
 (b  a )
 area 
R   dR 
Inside-out:

 (b  a
2
2
)
L
0
dz 
L
 (b 2  a 2 )
dr
dr
 length 


dR  d  



2 r L
A(r )
 area 
 b

b
R   dR 
dr

ln
 
2 L a
2 L  a 
Resistance of an object with arbitrary shape
dz
a
b
L
R   dR   
 
L
0

dL
dz
dz
 
  2
A
A( z )
 r ( z)
dz
  L  L du
 
z 0 2
2
 [a  (b  a) z / L]   b  a  u
  L  1 1   L

   
  b  a  a b   ab
Reduction of a resistive network
R
R
R
R
R
R

R
R
R
R

2R
R/2
R
2R

Reduction of a resistive network
R
R
b
R
R
b
a
a

R
R
a
R
b
R
R

R
R/3

(7/3)R
R
R
A freq. test question is what is the magnitude and sign of
the potential difference of a relative to b.
Reduction of a resistive network
Apply 42 V between a and c.
What is I between a and c?
I=3A
What is Vbc?
Vbc = 6 V
What is I2?
I2 = 2 A
Compare the brightness of
the four identical bulbs in
this circuit.
R
R
R
V
D is in parallel with a zero-resistance wire. The current will
take the zero-resistance path and bypass D altogether.
A and B are in series. So they will burn equally bright.
Together, they see the full battery voltage.
C experiences the full battery voltage, or twice the voltage
experienced by A or B. So C is much brighter.
If R1 is removed, R2 will glow
(1) more brightly.
(2) less brightly.
(3) same brightness as before.
If R1 is removed, R2 will glow
(1) more brightly.
(2) less brightly.
(3) same brightness as before.
Household devices are wired to run in parallel!
Strings of 50 Christmas lights in
series. Assume ~100 V source
and 25 W power consumption.
http://www.ciphersbyritter.com/RADELECT/LITES/XMSLITES.HTM
What is the resistance of a single bulb? (thought
question.)
A. 2
B. 4 C. 8
D. 10 
Gustav
Kirchhoff
•1824 – 1887
•German physicist
•Worked with Robert
Bunsen
•Kirchhoff and Bunsen
– Invented the spectroscope
and founded the science
of spectroscopy
– Discovered the elements
cesium and rubidium
– Invented astronomical
spectroscopy
Section 28.3
Multiloop circuits

Branch: An independent current path experiences only one current
at a given moment. It may be a simple wire or may also contain one
or more circuit elements connected in series.
Junction: A point where three or more circuit branches meet.
Loop: A current path that begins and ends at the same circuit point,
traversing one or more circuit branches, but without ever passing the
same point twice.
Multiloop circuits

The circuit above has: ? branches, ? junctions, ? loops.
3 Independent Equations.
To solve for 3 unknown branch currents, we need 3 equations.
To get these equations, use all but one (2  1 = 1) junction, and as
many independent loops as needed ( 3  1 = 2).
Kirchhoff’s current rule: I n  0
Current rule: The total current flowing into a junction is zero.
Arrows define positive branch-current directions. A current later
determined to be “negative” flows opposite its arrow.

 I  I1  I 2  0
 I  I1  I 2  0
 I  I1  I 2  0
Kirchoff’s Voltage Rule:  Vn  0
Voltage rule: The voltage changes around a loop sum to zero.
Arrows define positive branch-current directions.
V = + for a battery crossing from – to + terminal.
Use V = –I R when crossing a resistor in the positive direction.
Use V = +I R when crossing a resistor in the negative direction.
Alpine loop elevation
   I1R1  0

Single-loop circuit example
 20  I (2000)  30  I (1000)
 I (1500)  25  I (500)
 15  I (5000)  0
I
I  15 / 5000  3 mA
VA  Vground  20  I (2000)  30  I (1000)
 10  I (3000)  10  (0.003)(3000)
 19 V
Multiloop circuit example
Bottom loop:
I2 = I1
 5  I 2 (1)  I 3 (2)  0
Substitute I2 in junction Eq:
I3 = 2I1
 I 2  I1  I 3  0
Substitute I2 and I3 in top loop:
I1 = 5/3
Solve for currents:
I1 = 5/3, I2 = 5/3, I3 = 10/3
 5  I1 (1)  I 2 (1)  5  0
Multiloop circuit example
 6I1  10  4I 2  14  0
I 2  1.5I1  6
 2I 3  10  6I1  0
I 3  3I1  5
 I1  I 2  I 3  0
I1  (1.5I1  6)  (3I1  5)
 5.5I1  11  0
I1  2 I 2  3 I 3  1