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V  I R
voltage
All Scalar
quantities
resistance
current
Ohmic Conductor
If the p.d. across a length of wire is increased, the current
increases proportionally.
Slope = Resistance
Non-Ohmic Conductor
If the p.d. across a bulb is increased, it will get hot and it’s resistance
will therefore increase. The p.d. vs. current line will become steeper.
Non Linear
Resistance incerases
with temperature
V  I R
Not all materials follow Ohm’s law!
Those that do are called ohmic
Those that do not are called nonohmic
Ohmic
Nonohmic
Current and Voltage are linearly
proportional
circuits
Series circuits have all their components wired so that
current follows a single path through the circuit.
Have you ever hooked up strings of Christmas lights?
Remember how you plugged one end of a string into
another string, and so on? Then you have wired circuits
in series.
Series Circuits
Only one path for the electrical current (water).
Current flowing through
each element is the same.
IT  I1  I 2  .....
Total resistance equals the
sum of the individual resistors
RT  R1  R2  ...
Total voltage equals the sum
of the individual voltages
VT  V1  V2  ...
Series Circuits
If R1 = 3,000, R2 = 6,000 , and R3= 9,000 and the battery
supplies a voltage of 100 volts, Find:
a.
b.
c.
d.
e.
f.
IT
I1
I2
V1
V3
PT
Series Circuits
If the Emf of battery is 12 V, and the 3 resistors are identical, what
is the potential difference across each resistor?
a.
12 V
b.
0V
c.
3V
d.
4V
In the circuit R1 is 3 , R2 is 4 , and R3 is 2 . The ammeter
reading the total current leaving the battery records 1.50 A. Find:
1. The voltage dropped over each resistor.
2. The total resistance of the circuit?
3. The Potential difference of the battery, VT?
4. The power supplied by the battery?
A parallel circuit is a closed circuit,
in which the current divides into two
or more paths before recombining to
complete the circuit.
They are wired in such a way so that
if one part of the circuit is broken,
the whole circuit is still closed!
Many complex electronic devices are
wired in parallel. This allows a
single source to provide power to
many different components inside a
device, such as a stereo system.
Parallel Circuits - There’s more than one path for
the current to flow.
The total current is the sum
of the individual currents.
IT  I1  I 2  I3  ...
The voltage across any branch
is the same as the total
V=V=V=V
1
1
1
1
 
  ....
RT R1 R2 R3
The total resistance, RT is less
than the smallest one
Each new path reduces the resistance to the flow of water
The more paths for the
water (current) to flow the
less resistance there is to
the water (current) flow.
Homes are wired in
Parallel, WHY?
Power – The rate at which electric energy is being
used. Measured in Watts (scalar)
P  IV
V
V 
P   V 
R
R
2
P  I ( I  R)  I R
2
Current Divider Rule
 Current seeks the path of least resistance
 The
current entering any number of parallel resistors divides as the inverse ratio of their ohmic
value
Series-Parallel Circuits
Most circuits are actually a combination of both series
and parallel branches.
Find:
1. RT
2. IT
3. I1
4. I2
5. I3
6. V1
7. V2
8. V3
RT = R1 + R2 + R3
= 3 M + 3 M + 540 k
= 6.54 x 109 
VT
IT = VT / RT
= 42 V / 6.54 x 109 
= 6.42 x 10-9 A
VT
V1  I1  R1
 (6.42 106 A)(3 106 )
 19.3V
I1 = IT = 6.42 uA
I2 = IT = 6.42 uA
I3 = IT = 6.42 uA
V2  I 2  R2
 (6.42 106 A)(3 106 )
 19.3V
V3  I 3  R3
 (6.42 106 A)(540 103 )
 3.5V
RT = R1 + R2 + R3
= 880 k + 480 kW + 930 k
= 2.29 M
Find:
1.
2.
3.
4.
5.
6.
7.
8.
RT
IT
I1
I2
I3
V1
V2
V3
VT
IT = VT / RT
= 19 V / 2.29 x 106 
= 8.29 x 10-6 A
VT
I1 = IT = 8.29 mA
I2 = IT = 8.29 mA
I3 = IT = 8.29 mA
V1 = I1R1
= 8.29 x 10-6 A x (880 x 103) 
= 7.30 V
V2 = I2R2
= 8.29 mA x 480 k
= 3.98 V
V3 = I3R3
= 8.29 mA x 930 k
= 7.71 V
1.
2.
Kirchhoff’s Current Rule – Junction Rule
Kirchhoff’s Voltage Rule – Loop Rule
Kirchhoff's Current Law - The sum of the currents
entering any node (junction) equals the sum of the
currents leaving that node (junction).
Law of Conservation of Charge
Stated another way
Total current in = Total current out
Three ammeters are located near junction P in an electric circuit
as shown. If A1 reads 8.0 Amps and A2 reads 2.0 Amps, then the
reading of ammeter A3 could be:
a. 16 A
b. 6.0 A
c. 5.0 A
d. 4.0 A
As an analogy, consider vehicles at a road junction. The
number of vehicles passing point 1, per minute, must be equal
to the number of vehicles passing point 2 per minute plus the
number of vehicles passing point 3 per minute.
Total current into any junction =
Total current out of any junction
Total current into any junction =
Total current out of any junction
PARALLEL CIRCUIT
I  I1  I 2
Total current into any junction =
Total current out of any junction
I1  I 2  I 3
I 3  I1  I 2
Junction Rule
What is the current at point P?
a.
2A
b.
3A
c.
5A
d.
6A
e.
10 A
5A
P
8A
2A
Junction Rule
What is the current at point P?
a.
2A
b.
3A
c.
5A
d.
6A
e.
10 A
5A
P
8A
junction
2A
6A
Kirchhoff's 2nd Law - Loop Rule - The sum of the applied
potentials and the potential drops around
any closed circuit loop is zero.
V1  V2  V3  ....  0
Law of Conservation of Energy
In the equation below voltage drops represented with a – sign and
voltage gains with a + sign.
v1
I
v1
R1
- IR1
R2
-IR2
Move clockwise around circuit:
=0
Kirchhoff's Voltage Law - The algebraic sum of the voltages
around any closed path is zero.
B
A
B
C
A
When you go from – to + you gain voltage
When you go from + to – you loose voltage
Find the difference in potential between points:
A-B, B-C, C-A in both diagrams.
C
Loop Rule Example
Current in each resistor?
–IR = –8 V
I
–IR = – 4 V
Total resistance = 6 
Ohm’s Law: I = V / R
= 12 V / 6 
=2A
2
4
12 V
Start
+V = 12 V
Voltage drop across resistor?
V=IR
V = + 12 V – 8 V – 4 V = 0 