Ch. 20 – Electric Circuits
20 1 – Electromotive Force
20.1
Every electronic device depends on circuits.
Electrical energy is transferred from a power source
source, such as a
battery, to a device, say a light bulb.
Conducting
g
wires
+
Light bulb
A diagram of this circuit
ld llook
k lik
ll i
would
like th
the ffollowing:
Battery
+
-
Light
symbol
Battery
Inside a battery, a chemical reaction separates positive
and negative charges, creating a potential difference.
-
This potential difference is equivalent to the battery
s
battery’s
voltage, or emf (ε) (electromotive force).
This is not really a “force” but a potential.
Because of the emf of the battery, an electric field is produced within and parallel to the wires.
This creates a force on the charges in the wire and moves them around the circuit.
This flow of charge in a conductor is called electrical current (I).
A measure of the current is how much charge passes a certain point in a given time:
Electrical Current
Units?
I=
Δq
Δt
⎡ Charge ⎤ ⎡ C ⎤
⎢⎣ time ⎥⎦ = ⎢⎣ s ⎥⎦ = [Ampere] = [A ]
If the current only moves in one direction, like with batteries, it’s called Direct Current (DC).
If the current moves in both directions, like in your house, it’s called Alternating Current (AC).
Light bulb
I
+
I
Battery
-
Electric current is due to the fflow off moving
electrons, but we will use the positive
conventional current in the circuit diagrams.
So I shows the direction of “positive” charge
flow from high potential to low potential.
e
I
20.2 – Ohm’s Law
The flow of electric current is very analogous to the flow of water through a pipe:
The battery pushing the current acts like the water pump pushing the water.
The voltage of the battery is analogous to the pump pressure – the higher the pump
pressure, the
th faster
f t I can push
h the
th water
t through.
th
h Thus,
Th
the
th larger
l
my battery
b tt
voltage, the greater my current.
V ∝I
Let’s make this an equality:
V = IR
This is Ohm’s Law.
The proportionality constant, R, is called the electrical resistance.
Units?
V ⎡ Volts ⎤ ⎡ V ⎤
R= ⎢
= ⎢ ⎥ = [Ohm] = [Ω]
⎥
I ⎣ Amp ⎦ ⎣ A ⎦
Define Resistor: A component of an electrical circuit that offers
resistance to the flow of electric current.
Resistor,
R
Symbol for resistors:
+ -
Straight lines have
essentially zero resistance
ε=V
20.3 - Resistivity
The electrical resistance of a conductor depends on its shape:
-Longer wires have more resistance
-Fatter wires have less resistance
Length
Thus,
L
R∝ .
A
Let’s make this an equality:
L
R=ρ
A
Cross-sectional area
The proportionality constant, ρ, is the electrical resistivity.
Units?
A ⎡ Ω ⋅ m2 ⎤
ρ=R ⎢
⎥ → [Ω ⋅ m ]
L ⎣ m ⎦
Resistivity is an intrinsic property
of materials, like density:
Every piece of copper has the same
resistivity, but the resistance of any one
piece depends on its size and shape.
ρ, R
ρ, R
Clicker Question 20-1
A copper wire with a circular cross section has a resistance
R What would
R.
o ld the resistance of the wire
ire be if the radi
radius
s of
the wire was reduced by a factor of 2?
1. R
2. 2R
3. 4R
4. R/2
5. R/4
Temperature Dependence of Resistivity
The resistance of most materials changes
g with temperature.
p
For good conductors (metals) the resistance decreases with decreasing temperature.
R
Conductors
R
Insulators
T
T
For insulators (poor conductors) the resistance increases with decreasing temperature.
For many materials, we find that:
R = Ro [1 + α (T − To )]
R = Resistance at temperature T
Ro = Resistance at temperature To
α is the temperature
coefficient of resistivity
α >0
α <0
F metals
For
t l
For insulators
20.4 Electrical Power
Our standard definition of power is:
Power =
Work
.
Time
So what would electrical power be?
From the definition of potential:
Thus,
qV
P=
t
V=
W
⇒ W = qV
q
⇒ P = IV
We can write this different ways using Ohm’s Law, V=IR:
P = I 2R
V2
P=
R
*So we have 3 ways of calculating electrical power
depending on what other quantities are known.
Electrical Energy
Work and Energy have the same units (Joules)
(Joules).
Thus
Thus,
Energy = Power × Time
Electrical companies, like Entergy,
measure y
your monthly
y energy
gy use this
way, in units of kilowatt⋅hours (kWh).
For example, if you used an average power of 1500 W for 31 days (744 hours),
your energy consumption would be:
E = (1.5 kW )(744 h ) = 1116 kWh
At a cost of roughly $0.13/kWh, this would be a monthly bill of $145.
1 kWh = 3.60 ×106 J
Example:
Two light bulbs, A and B, are connected to a 120-V outlet (a constant
voltage
lt
source).
) Light
Li ht b
bulb
lb A iis rated
t d att 60 W and
d lilight
ht b
bulb
lb B iis rated
t d
at 100 W. Which bulb draws the most current from the outlet?
1. A
2. B
P = IV
If P is bigger, than I will be bigger for
the same V.
20.5 Alternating Current
When the voltage
g provided
p
changes
g periodically
p
y with time!
V = V0 sin 2π ft
I 0 = V0 / R
I =V / R
Peak values
I = I 0 sin 2π ft
f = 60 Hz
frequency
Sinusoidal function
Power:
Mean Value of Power:
P = IV = I 0V0 sin 2 2π ft
1
P = I 0V0 ≡ I rmsVrms
2
I0
V0
I rms =
Vrms =
2
2
root-mean square value
20.6 Series Circuits
The first way is to wire them together in series:
The same current runs through two
components connected in series.
V1 and V2 are called voltage drops.
*We speak of currents running through resistors, and voltages drops across resistors.
Thus, the current through resistor R1 is I, and the voltage drop across R1 is V1.
How would we find the net resistance (equivalent resistance, Req) for resistors
connected in series?
For resistors connected in
series, the sum of the voltage
drops across all the resistors
must equal the battery voltage
voltage.
Th
Thus,
V = V1 + V2
B t ffrom Oh
But
Ohm’s
’ Law:
L
IReq = IR1 + IR2 ⇒ Req = R1 + R2
Thus,, for resistors wired up
p in series,, the equivalent
q
resistance is:
Req = R1 + R2 + R3 + ⋅ ⋅ ⋅
i.e. you just add them!!!
Clicker Question 20-4
What is the current I in the following
g circuit?
4Ω
2Ω
12 V
1.
2
2.
3.
4.
0.50 A
2 00 A
2.00
10.0 A
60.0 A
20.7 Parallel Circuits
For resistors connected in parallel
parallel, the voltage
drop across each resistor is the same.
I1
The current through each might be different. It splits: I = I1 + I2.
I2
Thus,
I
From Ohm’s Law:
Thus,
V1 = V2 = V .
Req = V
V
=
=
I I +I
1
2
1
1
1
1
= +
+ + ⋅⋅⋅
Req R1 R2 R3
V
=
V1
V2
R1 + R2
1
1
1
+
R1
R2
for resistors in parallel.
Clicker Question 20- 5
Given the circuit below, a larger current flows through
which
hi h resistor?
i t ?
=2 Ω
=4 Ω
I
+V = 12 V
1.
2.
3.
R1
R2
The current through
each is the same.
20.8 Series and Parallel Circuits
Now let’s hook resistors up both in series and in parallel in the same circuit!
What is the current I in the following circuit?
I
I
I
I
We need to find the
equivalent resistance!
Thus,,
I
I
Req = 240 Ω
V
24
I=
=
= 0.1 A
Req 240
Clicker Question 20- 6
What is the equivalent resistance between points
A and
d B?
8.00 Ω
1.
2.
3.
4.
2Ω
4Ω
6Ω
8Ω
Three identical light bulbs are connected to a battery as
shown in the diagram.
g
What happens
pp
when bulb #3 burns
out?
1.
2.
3.
4.
Bulbs 1 and 2 both get brighter
Bulb 1 gets brighter and 2 stays
the same
Bulb 1 gets dimmer and 2 gets
b i ht
brighter
Bulb 1 gets dimmer and 2 stays
the same
#2
#1
#3
+-
C1
QTot = Q1 + Q2 = C1V + C2V = (C1 + C2 )V = CeqV
C2
Thus,
Ceq = C1 + C2 + C3 + ⋅ ⋅ ⋅
+ -
V
To find the equivalent capacitance for
capacitors wired in parallel, you just add them!
Now consider two capacitors wired in series:
C1
C2
V1
V2
The voltage across each will now, in general, be different,
but the
t ec
charge
a ge o
on eac
each o
of the
t ep
plates
ates must
ust be tthe
e sa
same.
e
V = V1 + V2 =
+ -
Q
C1
V
Thus,
1
Ceq
+ CQ2 = Q
(
1
C1
⇒Q=
=
1
C1
+
1
C2
+
1
C3
*Notice that these rules are just the
opposite for resistors in combination.
+ ⋅⋅⋅
+ C12
(
1
C1
+
)
)
1 −11
C2
V = CeqV
For capacitors wired in series, the
reciprocal of the equivalent
capacitance is just the sum of the
reciprocal capacitances.
Clicker Question 20- 7
What is the equivalent capacitance between points A and B?
1.
2
2.
3.
4.
0.5 μF
1 2 μF
1.2
2.0 μF
6 μF
Clicker Question 20-7 (cont.)
1
Let’s first number the capacitors:
2
3
4
Capacitors 2, 4, and 6 are in series, thus:
1
6
5
C 246
=
+ C14 + C16 =
1
C2
1
24
+ 121 + 18 =
1
4
⇒ C246 = 4.0 μF
1
Now the circuit looks like this:
246
3
4.0 μ
μF
5
Now capacitor 3 is in parallel with capacitor 246:
C2346 = C3 + C246 = 4 + 4 = 8
1
2346
Now the circuit looks like this:
8.0 μF
Now capacitors 1,2346, and 5
are in series:
5
1
C123456
=
1
C1
1
+ C2346
+ C15 = 15 + 18 + 16 ≈
1
2
⇒ Ceqq = C123456 = 2.0 μF