Download Week 6 - Circuits, Power and the Electromotive

Week 6 - Circuits, Power and the Electromotive Force
Exercise 6.1: Circuits
What are the voltmeter and the ammeter readings in each case?
Figure 1: Circuit diagrams
Answer: (a):
I = 2A
Vcd = 10V
(b):
I = 2A
Vcd = 10V
(c):
I = 0A
Vcd = 12V
Solution: (a):
I=
E
12V
=
= 2A
R+r
(5 + 1)Ω
Vcd = E − Ir = 10V
(b):
I = 2A
1
Vcd = IR = 10V
(c):
I = 0A
Vcd = 12V
Exercise 6.2: Discussion Questions
a) Two copper wires with different diameters are joined end to end. If a current flows in the wire
combination, what happens to electrons when they move from the larger-diameter wire into the
smaller-diameter wire? Does their drift speed increase, decrease or stay the same? Explain your
reasoning.
Answer:
The current can’t change in the diffrent parts of the wire. Therefore the drift speed has to increase
in the smaller diameter wire in order to keep the same amount of charge flowing per unit time. This
is similar to water flowing in pipes with different diameter. The speed of the water increases as the
diameter decreases to keep the same amount of water flowing in the whole pipe.
b) Batteries are always labeled with their emf; for instance, an AA flashlight battery is labeled ’1.5 volts.’
Would it also be appropriate to put a label on batteries stating how much current they provide? Why
or why not?
Answer:
No it would not. Because how much current depends on the kind of circuit the battery is connected
to. In the case of being used in a flash light, it depends on the luminous resistor in the light bulbe.
Different resistance implies different current.
c) Why does an electric light bulb nearly alwas burn out just as you turn on the light and almost never
while it is shining?
Answer:
When the lightbulb has been shining for a while, it’s temerature has more or less stabalized which
means that it’s resistance has stabilized. This value of the resistance is higher than the value of the
resistance when the light is just turned on. In fact the lowest resistance occurs at the moment when
you turn on the light and therefore also the highest current.
It’s the high current that is responsible for this burn out. What happens is that the sudden high
current of electrons causes the filament in the bulb to heat up and expand. The expansion is greatest
in the beginning and this can break an old filament.
Exercise 6.3: Transmission Line
A copper transmission cable 100 km long and 10.0 cm in diameter carries a curent of 125 A. The resistivity
of copper is ρcopper = 1.72 × 10−8 .
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a) What is the potential drop across the cable?
Solution: The potential drop is given by the familiar Ohm’s law V = IR, but in order to use this
we need the resistance R. This is related to the resistivity, the length and the diameter by
R=ρ
L
L
4L
=ρ
= ρ 2 = 0.22 Ω
d 2
A
πd
π 2
(1)
Answer:
V = IR = 27.4 V
(2)
b) How much electrical energy is dissipated as thermal energy every hour?
Solution: The power dissipated in the transmission cable is given by
P = IV =
V2
= 3423 W
R
(3)
and since this is the energy per unit second and there is 60 × 60 = 3600 seconds in an hour we get
that the energy per unit hour is given by
3423 × 3600 J/h ≈ 12 MJ/h
(4)
Exercise 6.4: Motion of particle in electric field
We are going to write a program that calculates the path of a particle that moves in an electric field
E. For simplicity we will do the time-integration using Euler-Cromer’s method. In this method, we
calculate a position r(t) at time ti+1 discretely as follows:
v(ti+1 ) = v(t) + a · dt
(5)
r(ti+1 ) = r(t) + v(ti+1 ) · dt
(6)
We are from here on and out working only with natural units. This means that all our variables are
given as dimensionless, making it much easier to work with them numerically.
The particle has mass m = 3.4 and charge q = 2.2. It starts out in the position r(t0 ) = [1.8, 1.2, 10.0]
and with velocity v(t0 ) = [3.4, 0.0, 0.0]. The electric field is E = [0.0, 0.0, −0.2].
a) Calculate the motion of the particle from t = 0 to t = 10 with time step dt = 10−3 . Plot the particle’s
x and z position as a function of t.
b) Find the analytical solution and compare it with the numerical solution by plotting it.
We will now vary the electric field with time. Define E(t) = −0.2 · 1 − e−t/5.2 k̂.
c) Use the same program as you used above to calculate the new path of the particle. Make plots for x
and z as functions of t.
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d) Make a plot showing the new path in an x, z-coordinate system.
Answer:
10
Constant field
Varying field
9
8
z
7
6
5
4
3
2
0
2
4
t
6
8
10
Figure 2: This plot only shows the z-coordinate as function of time.
Example program that plots the field, but only plots the z position of the particle with constant and
varying field.
from p y l a b i m p o r t ∗
r c ( ’ t e x t ’ , u s e t e x=True )
d t = 1 e−3
t0 = 0
t 1 = 10
t = l i n s p a c e ( t0 , t1 , ( t 1 − t 0 ) / d t )
E = 2 . 2 ∗ ( 1 − e x p (− t / 5 . 2 ) )
figure ()
p l o t ( t , E)
x l a b e l ( " $ t $ " , f o n t s i z e =16)
y l a b e l ( " $E$ " , f o n t s i z e =16)
s a v e f i g ( " f i e l d . pdf " )
q = 2.2
m = 3.4
r1 = zeros (( len ( t ) ,3) )
v1 = z e r o s ( ( l e n ( t ) , 3 ) )
r2 = zeros (( len ( t ) ,3) )
v2 = z e r o s ( ( l e n ( t ) , 3 ) )
r1 [ 0 ] = [ 1 . 8 , 1.2 , 1 0 . 0 ]
v1 [ 0 ] = [ 3 . 4 , 0 . 0 , 0 . 0 ]
r2 [ 0 ] = [ 1 . 8 , 1.2 , 1 0 . 0 ]
v2 [ 0 ] = [ 3 . 4 , 0 . 0 , 0 . 0 ]
for i in
E1 =
a1 =
E2 =
a2 =
r a n g e ( l e n ( t ) −1) :
a r r a y ( [ 0 . 0 , 0.0 , −2.2])
q∗E1/m
a r r a y ( [ 0 . 0 , 0 . 0 , −2.2∗(1 − e x p (− t [ i ] / 5 . 2 ) ) ] )
q∗E2/m
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v1 [
r1 [
v2 [
r2 [
i +1]
i +1]
i +1]
i +1]
=
=
=
=
v1 [
r1 [
v2 [
r2 [
i
i
i
i
]
]
]
]
+
+
+
+
a1 ∗1 e−4
v1 [ i +1]∗ d t
a2 ∗1 e−4
v2 [ i +1]∗ d t
figure ()
p l o t ( t , r 1 [ : , 2 ] , l a b e l=" C o n s t a n t f i e l d " )
h o l d ( ’ on ’ )
p l o t ( t , r 2 [ : , 2 ] , ’−− ’ , l a b e l=" V a r y i n g f i e l d " )
x l a b e l ( " $ t $ " , f o n t s i z e =16)
y l a b e l ( " $z$ " , f o n t s i z e =16)
legend ()
s a v e f i g ( " motion . pdf " )
show ( )
Exercise 6.5: Measuring on a Circuit
Figure 3
Figure 3 shows a circuit where a voltmeter V is attached across the terminals of a battery. Here the
resistance indicated between the measuring points of the voltmeter is the internal resistance r of the
battery. This battery is connected to a circuit with a resistance R, a switch S and an ammeter A.
Remember that no current flows trough the voltmeter and that there’s no voltage drop (or ressistance)
associated with the ammeter.
When switch S in figure 3 is open, the voltmeter V connected across the battery reads 3.5 V. When the
switch is closed, the voltmeter reading drops to 3.0 V, and the ammeter A reads 1.5 A. Find the emf,
the internal resistance of the battery, and the circuit resistance R.
Solution:
When no current is running, the voltage measured is equal to the emf, so ε = 3.5 V. When the current
is turned on the terminal voltage is equal to the voltage drop across the resistor R. We have that
ε − Ir = 3.0 V ⇒ r =
ε − 3.0 V
= 0.33 Ω
I
(7)
and we also know that the voltage across the terminal is 3.0 V so that
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ε − Ir = IR ⇒ R =
ε − Ir
= 2.0 Ω.
I
(8)
Answer:
ε = 3.5 V
(9)
r = 0.33 Ω
(10)
R = 2.0 Ω
(11)
Exercise 6.6: Lightning Strike
A lightning bolt strikes one end of a steel lightning rod, producing a 30000 A current burst that last for
65 µs. The rod is 1 m long and 2 cm in diameter, and its other end is connected to the ground by 40 m
of 5.0 mm-diameter copper wire. The resistivity of steel is given by ρsteel = 20 × 10−8 Ωm and copper is
ρcopper = 1.72 × 10−8 Ωm.
a) Find the potential difference between the top of the steel rod and the lower end of the copper wire
during the current burst.
Solution:
Let the length of and resistivity of the rod and wire be L, ρsteel and l, ρcopper respectively. Then, for
a rod and a wire, the resistance is related to it’s length, diameter and resistivity by
R=
ρL
4ρL
=
.
A
πd2
(12)
So the total resistance of the rod and wire is therefore
4
R=
π
ρsteel L ρcopper l
+
D2
d2
4
=
π
20 × 10−8 × 1
(2.0 × 10−2 )
2
Ω+
1.72 × 10−8 × 40
2
(5.0 × 10−3 )
!
Ω
= 0.0356 Ω.
(13)
So the potential difference between the top of the steel rod and the end of the wire is
V = RI = 0.0356 Ω × 30000 A = 1.07 kV
(14)
V = 1.07 kV
(15)
Answer:
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b) Find the total energy deposited in the rod and wire by the current burst.
Solution:
The power or energy deposited per second is
P = V I = 1.07 kV × 30000 A = 3.2 × 107 W = 32 MW
(16)
so that the total energ deposited is
32 MW × 65 µs = 2.09 kJ
(17)
A little bit suprised by the low energy dissipation. Clearly this is because of the low resistivity of steel
and copper. Constrast it with the resistivity of semiconductors and insulators. There the resisivity is
10 to 23 order of magniture higher, which would imply an immensely difference in power dissipation.
It does however make sense that one would not want the lightning to dissipate all of the energy in
the rod, when this would tear it apart. Now the energy is dissipated safely in the ground instead.
Answer:
Edeposited = 2.09 kJ
(18)
Exercise 6.7: Maximal Power Output
A source with emf ε and internal resistance r is connected to an external circuit.
a) Show that the power output of the source is maximum when the current in the circuit is one-half the
short circuit current of the source.1
Solution:
For any source the power output is given as
P = IV = εI − I 2 r
(19)
where V is the potential difference over the source while I is the current running trough it. What is
the short circuit current? It is the current running trough the source, when the source is connected
only to itself. I.e. it is given by
Isc = rε .
The maximal power output of the source happens when
dP
d
=0⇒
εI − I 2 r = 0
dI
dI
1 The
(20)
short circuit current is the current you get when the source is connected to itself.
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or equivalently
I=
1ε
1
= Isc .
2r
2
(21)
b) If the external circuit consists of a resistor R (also called a load2 ), show that the power output of the
source is maximum when the internal resistance r of the source equals the load resistance R. Express
Pmax in terms of the EMF ε and the internal resistance r.
Solution:
If the external cicuit consists of a load resistance R the total current trough it is given by
V = IRtot = I(r + R) ⇒ I =
V
r+R
(22)
so the power output of the load is then
P = IV = I 2 R =
V
r+R
2
R.
(23)
Differentating this expression with respect to R we get a maximum when r = R. Now this expression
is also maximised with respect to the current when the source delivers it’s maximal current. Therefore
we get a power output of
Pmax = I 2 R =
ε 2
ε2
r= .
2r
4r
(24)
Answer:
Pmax =
2A
ε2
4r
(25)
resistor or ’load’ may represent your toaster, hair dryer or any electrical equipment which dissipate heat.
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