and
Work
Power
IV
Now recall that q/t is defined to be a
current I so:
Power
qV
t
Putting these two together we find:
Power
Work
time
qV
Recall the definition of Power and the
work done in moving a charged particle
across a potential difference:
Electric Power
V
R
I
IV
Power
V V
R
or if you use I
V
R
2
V
R
Now recall that V = I R so:
2
Power I I R I R
Power
Electric Power Definition
There is a second type of current called
alternating current (ac) in which the emf
which "drives" the current switches
direction at some angular frequency. A
generator produces this type of current.
The reason this occurs will be explained in
lecture.
Up until now we have been thinking about
charges which move through a circuit in
one direction. This is called a direct current
(dc) circuit. A battery produces this type of
current.
AC Electricity
V
R
Recall:
V0
sin 2
R
V 0 sin 2
ft
ft
I
I 0 sin 2
ft
V0
but
is defined as I 0 so:
R
I
V
AC Electricity
2
I 0 V 0 sin 2
1
I0V0
2
ft
ft
1
1 1
so:
but you may write as
2
2 2
I0 V0
P
I rms V rms
2
2
The average power is given by: P
therefore: P
P
Recall: P I V so:
I 0 sin 2 f t V 0 sin 2
AC Power
Ohm's Law then becomes
V rms I rms R
V
V
R
R
R
I
I
What do you do for
more complicated
circuits? The circuit on
the left is an example
of two resistors in
series.
Ohm's Law will tell us how to
solve an unknown part of the
circuit. Given V and R, you can
solve for I etc.
Circuits
V
V
R1
R1
R2
R2
Resistors in Parallel
Resistors in Series
Series and Parallel Resistors
I
VT V1 V2
I RT I R1 I R2
RT R1 R2
Because of this, you can simple add the
two resistors together to find the total
resistance. The reason is shown below:
Since there is only one path the current
may travel, there is one single I through
each resistor.
Series Resistors
V
I1
R1
I2
R2
V
R1
1
R1
1
RT
1
R2
V
R2
I1 I2
V
RT
IT
Here, the voltage across each resistor is
the same but the current is split between
the two resistors.
Parallel Resistors
V
R1
R2
R3
therefore:
2
3.33 10
1
30
R2,3
R2 and R3 are in parallel!
1
1
1
R2,3 75
50
1
2
3.33 10
R2,3
Find the current in the following circuit and power
dissipated in the resistor if: V = 12V, R1 = 100Ω, R2
= 75Ω, and R3 = 50Ω.
Example Problem
I
I
V
R1,2,3
therefore:
12V
9.2 10
130
R1,2,3
R2,3
R1 and R2,3 are in series!
R1,2,3 R1 R2,3
R1,2,3 100
30
therefore:
R1,2,3 130
V
V
R1
Example Problem (2)
2
A
V
P
V2
R
12V 2
130
144 V 2
130
1.11 Watts
The power dissipated across the reisistor is given by:
V2
P I V or P
or P I 2 R
R
Any of them may be used.
R1,2,3
Example Problem (3)
This resistance is caused by the
chemicals internal to the battery
having a measurable resistance.
We have only considered ideal
batteries up to this point. This
means we have only considered
batteries to be sources of
voltage. In reality, batteries
themselves has a small
resistance associated with them
which needs to be taken into
account.
Internal Resistance
V
r
R1
R2
R3
You would solve the problem as you
did before but in the end, calculate a
total resistance Rr,1,2,3 with r and R1,2,3
in series with one another.
The circuit used in the previous
example should have shown one
extra resistance associated with the
internal resistance of the battery.
Correct Circuit