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 for the case in figure 1 (a), (b) and (c) respectively?
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
Figure 2
Exercise 6.2:
Consider the circuit shown in figure 9. The current trough the 6.00 Ω resistor is 4.00 A, in the direction
shown. What are the currents trough the 25.0 Ω and 20.0 Ω resistors?
Solution: Let’s call the resistor with X Ohms for RX and the current trough it IX . In order to find the
current trough R25 we must find the current trough R8 . This can be found by the loop rule applied to
the loop containing both R6 and R8 . We get
I6 R6 = 4.00 × 6.00 AΩ = I8 R8 = I8 8.00 Ω
(1)
implying
I8 =
4.00 × 6.00
A = 3.0 A.
8.00
(2)
The current trough R25 is then just
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I25 = I8 + I6 = (4.00 + 3.00) A = 7.00 A.
(3)
In order to find the current trough R20 we apply the loop rule trough the larger loop containing R20 ,
and the equivalent resistance Req of the three resistors R6 , R8 and R25 . We get
I25 Req = I20 R20 .
(4)
R6 R8
+ R25 = 28.4 Ω
R6 + R8
(5)
I25 Req
7.00 × 28.4
A = 9.94 A.
=
R20
20.0
(6)
where
Req =
implying a current of
I20 =
Answer:
I25 = 7.00 A
(7)
I20 = 9.94 A
(8)
Figure 3
Exercise 6.3:
In the circuit shown in figure 10, find (a) the current in the 3.00 Ω resistor; (b) the unknown emfs ε1
and ε2 ; (c) the resistance R. Note that three currents are given.
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Solution:
Figure 4 shows the assumed direction for the different currents in the circuit. Let’s use the same
convention as in the last exercise and label the resistance with X Ohms and the current going trough it
as RX and IX . From the junction rule we can write down that
IR
=
I4 + Iε1
(9)
Iε1
=
Iε2 + I3
(10)
Iε3
=
I6 + IR .
(11)
Substituting 11 into 10 and we get
Iε1 = I6 + IR + I3 .
(12)
We can further substitute this result into 9 and obtain
IR = I4 + I6 + IR + I3 ⇒ I3 = −(I4 + I6 ) = −8.00 A
(13)
which means that the current is going in the opposite direction relative to the assumed one.
To find the unknown emfs we apply the loop rule and obtain
ε2 = −I3 R3 + I6 R6 = (8.00 × 3.00 + 5.00 × 6.00) V = 54.0 V
(14)
ε1 = −I3 R3 + I4 R4 = (8.00 × 3.00 + 3.00 × 4.00) V = 36.0 V.
(15)
Applying the looprule to the uppermost rule we get
IR R = ε2 − ε1 ⇒ R =
ε2 − ε1
54.0 − 36.0
Ω = 9.0 Ω.
=
IR
2.00
(16)
Answer:
IX is the current trough the resistor with X Ohms. Note that the sign of the current depends on which
direction one assumes it is going.
Week 6 – September 25, 2012
I3 = −8.00 A
(17)
ε1 = 36.0 V
(18)
ε2 = 54.0 V
(19)
R = 9.0 Ω
(20)
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Figure 4
Exercise 6.4: 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 different 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.
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Exercise 6.5: Measuring on a Circuit
Figure 5
Figure 5 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 5 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
(21)
and we also know that the voltage across the terminal is 3.0 V so that
ε − Ir = IR ⇒ R =
ε − Ir
= 2.0 Ω.
I
(22)
Answer:
Week 6 – September 25, 2012
ε = 3.5 V
(23)
r = 0.33 Ω
(24)
R = 2.0 Ω
(25)
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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
(26)
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
(2.0 × 10−2 )
Ω+
1.72 × 10−8 × 40
(5.0 × 10−3 )
2
!
Ω
= 0.0356 Ω.
(27)
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
(28)
V = 1.07 kV
(29)
Answer:
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
(30)
so that the total energ deposited is
32 MW × 65 µs = 2.09 kJ
(31)
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
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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
(32)
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
(33)
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
(34)
or equivalently
I=
1ε
1
= Isc .
2r
2
(35)
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
(36)
1 The
2A
short circuit current is the current you get when the source is connected to itself.
resistor or ’load’ may represent your toaster, hair dryer or any electrical equipment which dissipate heat.
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so the power output of the load is then
P = IV = I 2 R =
V
r+R
2
R.
(37)
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
(38)
Answer:
Pmax =
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ε2
4r
(39)
compiled September 25, 2013