Download Mr. Sarver Review Questions 16 The current will flow only as long as

Mr. Sarver
Review Questions
16
The current will flow only as long as there is a potiential difference (gradient in calculus)
between the two ends of the conductor.
18
Electrons are free to move around. Protons are bound to the nucleus of the atom by the
strong force which are locked in place by the atomic lattice.
20
AC is alternating current. In this case, current moves forward then backward in a sinusoidal
pattern. DC is direct current. It flows in one direction and one direction only.
22
The greater resistance belongs to the thin wire.
24
Doubling voltage on constant resistance gets doubled current.
Doubling voltage and resistance gets the same current.
26
Dry skin has higher resistance as the ions can move around allowing electrons to move around.
28
The third prong of a plug has at least one use, it keeps the body of the device at ground
voltage ensuring that should anything go wrong on the inside of the device, that current is
directed to ground and not you. Should the device actually short to ground, the excessive
current will cause the break to pop.
The second use is for electronic devices. Electronics such as computers and TVs use it to
provide a constant zero voltage on shielding (Faraday cages) for certain sensitive parts.
30
Ann electric circuit is a loop of conductor that allows current to flow around it. A gap in
the circuit means it is no longer circuit and current no longer flows.
32
The potiental across the second bulb would be 4 V, Kirckoff’s Law.
34
It tells me that the resistance in the two branches of the circuit are equal.
36
As more parallel branches are closed in a parallel circuit the more current that can pushed
through the circuit at equal voltage.
1
38
The more devices that placed in parallel, the more current that drawn. The more current
drawn, the hotter the fuse gets until the fuse toasts and opens the circuit.
40
The 100 W bulb will draw more current.
Exercises
18
Increasing the current increases the brightness of the lightbulb.
22
5 is the only circuit the lights the bulb. 1 and 2 are closed circuits but do not light the bulb.
3 and 4 are open circuits.
24
The applicance uses energy carried by the electricity. That energy gets used by the applicance in some useful manner and then lost to space by dissipative means.
26
It is unfortuate that the book said the only a small amount of the energy gets turned into
light. It should have said that a small amount gets turned into visible light. The portion
that it is refering to that is fair amount of the energy in question is actually turned into
invisible light, most of it in IR and UV ranges, unless it was talking about flourecent bulbs.
That waste energy gets turned into invisible UV light at the bulb and invisible IR light at
the transformer. That buzz is the transformer in the fixture vibrating at 60Hz and is also
a loss of energy. LEDs loss energy to chrystal lattice vibrations called phonons or sound
waves, they are either not as loud as the transformer or outside the human hearing range.
30
Depends on the safety features and cost of the device, socket, and building. In a properly
wired building and 110V socket, plugging the 220V device into the 110V socket will cause the
breaker to pop and nothing else will happen. If the building is not up to code, the current
could melt the insulation, short out and burn the building to ground losing both building
and device. In that second case, plugging the 110V device into the 220V socket will probably
only cause the loss of the device unless it is protected against over-current/over-potential. I
dont think you could pull either trick in a properly wired building due to the design of the
sockets, so don’t worry about it too much.
In summary, plugging the 220V into a 110V socket will cause less damage by popping a
breaker rather than losing the device.
I sent this question to my dad to see what he thought of it. This is what he had to say (this
is the highly practical look and may be significantly more information then you should like):
Well, in some ways there is insufficient information but I’ll give it a go.
2
First, if the building is properly wired with the applicable outlets and the machines have the
appropriate plugs on them, then the plugs will not match the outlets in any case.
Ok, so let’s say we cut the plugs off and strip the wires and then remove the outlets and pull
the wires out where we can get at them.
1. Taking the 110V machine and hardwiring to the 220V circuit. The hot wire from the
machine will be tied to one of the two hot legs provided from the circuit, the neutral
to the neutral and the ground to the ground. I’m not real sure if a 220V circuit has
both a ground and a neutral. If not, then the neutral and ground from the machine
goes to the neutral or ground provided. The second hot wire from the circuit will have
to be capped, taped, and otherwise insulated to protect surroundings and people, etc.
When the power is supplied to the circuit, the machine will run very well at least for a
while. There will be no load protection for the machine and the motor will eventually
fail.
2. Taking the 220V machine and hardwiring to the 110V circuit. At first glance I don’t
know how you would do it practically.
• Suppose you secure one of the hot wires from the 220V machine and the neutral
and ground (if they are both there) to the neutral and ground of the 110V circuit.
Then when the power is supplied to the circuit and assuming the proper amperage
is available, the motor will ”single phase”. A220V motor runs with two phases
of electricity. Single phasing is using only one side of the power. The motor will
not turn, it may hum for a little while, but eventually it will produce smoke and
melt the coatings on the windings and maybe even catch fire.
• Suppose you secure both of the hot wires from the 220V machine to the one hot
wire from the 110V circuit and of course the neutral and ground to their wires.
I’ve never seen this tried. Yes, the first thing that should happen is the breaker
should trip. Suppose that the breaker is big enough to ignore the amperage draw,
I expect the motor still will not turn because it is designed to operate on two
phases of electricity that are 180 degrees out of phase with each other. I assume
that the motor will have the same results as the previous situation.
I don’t know for sure what would happen back at the breaker panel should the breaker in
question not trip. Usually if the breaker doesn’t trip, that means that the power continues
to pass through the wires to the motor until it burns up. I suppose eventually the wires will
fuse together short circuiting the system causing the wires to heat up and burn through the
insulation causing catastrophic failure of varying proportions.
You are right, if everything is up to code, these situations could not happen. I don’t think
the question is concerned with code nor considers the breaker panel. In both cases, as a
minimum, you will have motor failure.
One other possibility is that the smaller 220V motors have optional wiring patterns allowing the 220V motor to be wired for 110V power. If the 220V motor is correctly wired for
110V, then the 220V motor will draw twice the amperage in the 110V configuration but the
electrical systems will be happy.
3
32
I would expect to find AC in the filament of the incandescent light bulb in my house. I
would expect to find DC in the filament of the incandescent light bulbs in my car.
34
V = IR ⇒ I = VR
2
P = IV = I 2 R = VR
Insufficent information for me to tell. I need to know if the current through or the potiential
across the bulbs are the same. If you assert one or the other be sure to state your assumption. I suspect that the voltage is the same between the two in which case, the resistance is
less for the high-beams.
2
= 2.88Ω
RH = 144V
50W
12V
IH = 2.88Ω
= 4.17A
40W
RL1 = (4.17A)
2 = 2.304Ω
10W
RS = (4.17A)2 = .576Ω
2
RL2 = 144V
= 3.6Ω
40W
The resistance of the high beams in 12V circuit is 2.88 ohms. To take the same current, the
resistance of the bulb would have to be 2.304 ohms. The current will only be the same if a
.567 ohm, 10 watt resistor (heat Sink) is included in the circuit in series with the light bulb.
I think it would be easier to use the same potiental and a 3.6 ohm bulb for the low beam.
Approxamently 12V is the standard potiental in a car circuit.
38
I disagree. A real battery provides neither constant current nor constant potiental. The
battery can only provide so much current and that current is what determines the voltage.
That isn’t to say that the battery will do its best to provide a constant potiental. An ideal
battery will provide a constant potiential, but a real one will not.
Problems
9
I = VR ⇒ R = VI
R = 120V
= 6Ω
20A
10
P = IV ⇒ I = VP =
R = 120V
= 12Ω
10A
1200W
120V
= 10A
12
I1 + I2 = IT = 3A + 3A = 6A
60Ahr
= 10hr
6A
It will take longer to run it dead according to the lights. As the energy is pulled out, the
voltage drops. As the voltage drops, the current drawn drops. It will take less time to run
it dead according to the starter as the starter requires far more current than lights do.
4
13
$.2
kW h
×
kW
1000W
× 100W × 1we ×
7d×24h
we×d
= $3.36
5