Exercise 5A, Case 1

I can see I will need the following equations to help me solve the next few cases involving the power plant.
V=IR P=IV P=I2R
Where V represents Voltage drop, I represents Current, R represents Resistance, & P represents Power lost.

This intro tells me that I will be using a plant voltage of 120V to solve the first cases. I can see that I’ll be predicting various voltage drops across a completed circuit, from a power plant to a house.

Exercise 5A, Case 2

If nothing is hooked up in the house:
1.) What is the current through the supply wire?
To solve this I go back to my equation, deciding to use I=V/R , but thinking back on the intro I see there is nothing hooked up in the house; therefore, there is no current. The answer is 0.
2.) What is the voltage drop along the supply wire?
To solve you would use the equation V=IR, I is 0; therefore, the voltage drop will also be 0.
3.) What is the amount of power converted to thermal energy (Power loss)?
To solve this one, P=IV, I and V are both 0; therefore no power is lost in this Case.

My predicted answers were correct in this case. If there is nothing in the house, then no current is flowing through the wires.

Exercise 5A, Case 3

Now I’m asked that if I’m using super conducting wires with 0 electrical resistance and we plug in a 100 Watt lamp inside the house:
1.) What is the current through the circuit?
I=P/V (100 watts)/(120 volts)= 0.83 Amps
2.) What is the resistance of the lamp?
R=V/I (120 volts)/(0.83 amps)= 144.6 Ohms
3.) What is the voltage drop across the lamp?
120 volts
4.) What is the voltage drop across the supply & return wires?
There wouldn’t be a voltage drop along the wires because they have zero resistance. The voltage drop would only occur at the lamp.

My predicted answers were correct in this case also.

Exercise 5A, Case 4

I’m given the following information for this case: Resistance is .001 Ohm/meter, assume each wire is 1 km, so R=1 Ohm, with the 100 watt lamp:
1.) What is the voltage drop along each wire?
So I see I’ll be using the same current as in Case 3, which is .83 Amps.
Supply Wire: V=IR (.83 Amp)(1 Ohm)=.83 Volts
At House: V= 120 Volts- .83 Volts= 119.2 Volts
Return Wire: V=IR (.83)(1)= .83 Volts
2.) The power lost?
P=IV P= (.83 Amps) (.83 Volts) = 0.6889 Watts

The voltage drop at the house I got incorrect, but now I see I forget to count the voltage drop across the return wire as well. Therefore, the voltage drop is (.83 Volts)(2)= 1.66 Volts; so the voltage drop at the house is 120 V- 1.66 V= 118.3 Volts.

Exercise 5A, Case 5

Now two lamps are hooked up, so twice as much current is passing through the wires:
1.) What is the current through each wire?
I= (.83 Amps) (2) =1.66 Amps
2.) What is the voltage drop on the wires?
V=IR V=(1.66 Amps)(1 Ohm)= 1.66 Volts
P=I2R P=(1.66 Amps)2(1 Ohm)= 2.76 Watts
3.) What is the voltage drop on the house?
V={120 Volts- (1.66 Volts) (2)}= 120 Volts – 3.32 Volts= 116.7 Volts

No surprises in this case, we expect to see a greater power loss than in Case 4 because a greater current causes a larger voltage drop in the wires. This means that the house is using more electricity, which is explained by using two lamps instead of one.

Exercise 5A, Case 6

Now suppose our house is 300 km from the power plant. The wire is still R=.001 Ohm/Meter and our house is now drawing 10 Amps of current:
1.) What would be the voltage drop in each wire?
First I have to calculate the Resistance: R=(.001)(300000)= 300 Ohms
V=IR V=(10 Amps)(300 Ohms)= 3000 Volts
2.) What would be the power loss in each wire?
P=IV P=(10 Amps)(3000 Volts)= 30,000 Watts
3.) What voltage would the power plant have to produce for the voltage at the house to be 120 V?
(3000 Volts)(2)= 6000 Volts +120 Volts= 6120 Volts is what the power plant needs to produce.

My predictions were correct; however, anyone can see this power plant would be extremely inefficient because so much power is lost.

Exercise 5A, Case 7

Now they will be delivering 1200 Watts to the house, but the voltage drop at the house is set at 120,000 Volts instead of 120 Volts:
1.) What is the current in the circuit?
I=P/V I= (1200 Watt)/(120000 Volts)= .01 Amps
2.) What is the voltage drop across each wire?
V= IR V= (.01 Amps)(300 Ohms)= 3 Volts The total voltage drop for both wires would be 6 Volts.
3.) What is the power loss in each wire?
P=IV P=(.01 Amps)(3 volts)= .03 Watts The total power loss would be .06 Watts
4.) What would the total plant voltage would need to be?
120,000 Volts + 6 Volts= 120,006 Volts

In this case, we see a much more efficient power plant than Case 6; however, using voltage at these levels can be very dangerous. It explains in the description of the answers that the power company uses high voltage is accompany with transformers to create both efficient and safe power delivery.

Exercise 5B, Case 1

In these cases we will be determining the dangers of electrocution, in the reading we learn that the most dangerous levels are 500-1000 Volts. If the skin is wet than anything over 100 Volts can be dangerous. Low voltage can be dangerous, but are harmless as long as they don’t pass through the heart:

100 Volts, from Right foot to Right arm

This is a low voltage and if the skin is dry it may hurt, but will not be lethal because it doesn’t pass through the heart.

Exercise 5B, Case 2

600 Volts, from Right foot to Right arm

This voltage would be lethal is it passed through the heart, but it’s only on the right side. So it would be very painful, but not lethal.

Exercise 5B, Case 3

100 Volts, from the Right side of head to the Left hand

Voltage would pass through heart, but is not high enough to be lethal. I would expect it to be painful.

Exercise 5B, Case 4

600 Volts from Right ear to Left hand

This would be lethal because it passes through the heart and would cause the heart to fibrillate and stop.

Exercise 5B, Case 5

100 Volts from Right hand to Left hand

Painful, but not lethal if quickly removed; even though this passes through the heart it is not a high voltage.

Exercise 5B, Case 6

600 Volts from Right hand to Left hand

This would be lethal, it passes through the heart and would cause the heart to fibrillate and stop.

Exercise 5B, Case 7

12 Volts from Right hand to Left hand

This would be quite safe, it is a low voltage and wouldn’t even be very painful.

Exercise 5C, Case 1

We will be moving a coil through a magnet and see how it affects a light bulb that is in a circuit with it.

If the coil is moved slowly through the magnet, what will happen to the bulb?

I predict that the light bulb will light up. The bulb did light up, but did not stay on.

Exercise 5C, Case 2

If the coil is moved twice as fast, what will happen to bulb and voltage?

I predict that the bulb will light up and stay on for a longer time than before, but not permanently. As long as the coil keeps moving, the bulb will stay on, this is because the magnetic field is changing rapidly.

Exercise 5C, Case 3

If the coil is moved half as fast as in case 1, what will happen to bulb and voltage?

I predict that the light will not come on and the voltage will change, just not quickly enough. The light did come on, but it was dimmer than in case one and it actually stayed on for longer intervals.

Exercise 5C, Case 4

If you leave the coil fixed and move the magnet at the same speed as case 1, what will happen to the bulb and voltage?

I predict that the results will be the same as those in Case 1, the bulb will flicker on and off. This was correct.

Exercise 5C, Case 5

Now the number of turns in the coil has doubled, move the coil the same as Case 1, what do you predict will happen?

I predict the light will stay on for a longer period of time than it did in Case 1. It actually has the same time dependence as the coil in Case 1.

Exercise 5D, Case 1

We are making a circuit with wires, coil, battery, and a compass.

If the battery is disconnected, how will the compass needle point?

I predict that the needle will point to North. This is correct because there is no circuit and Earth’s magnetic field will cause the needle to point to North.

Exercise 5D, Case 2

Now we connect battery and run 1 amp of current through the circuit? How will the compass needle point?

I predict that the needle will point west towards the coil. The needle does move slightly to the left pointing towards west, but because the Earth’s field is stronger it points slightly northwest.

Exercise 5D, Case 3

Now we add another battery, how will the compass needle point?

I predict that the needle will point even more west this time. The second battery doubles the voltage and causes the needle to point a little more west, but is still northwest.

Exercise 5D, Case 4

Now we have two batteries and we double the turns in the coil, how will the compass needle point?

I predict that the needle will finally point all the way to west. The results are actually the same as they were in Case 3 because the current is halved.

Exercise 5E, Case 1

I can see that I’m setting up the equipment that I will need in these exercises. First I have an iron core, then a primary coil, a secondary coil with half as many turns, power supply, volt meter, and an amp meter.
Exercise 5E, Case 2

The primary coil has twice as many turns as the secondary and is connected to 100 V DC. What will be the voltage out of the secondary vs time?

I predict that there will not be a voltage output on the secondary coil because it is a direct current. I was right; you have to have a change or alternating current in order for the secondary coil to have voltage.

Exercise 5E, Case 3

Now the primary coil is hooked to a 100 V source that oscillates up and down 60 times per second. What will be the voltage out of the secondary vs time?

I predict that the voltage will be halved. The voltage is halved because there is half the number of turns on the coil.

Exercise 5E, Case 4

The number of turns on the secondary coil are increased by a factor of 5. What will be the voltage and current out of the secondary vs time?

I predict that the voltage will be increased by a factor of 5 and the current will increase also. The voltage did increase by a factor of 5 because that’s how many turns were increased on the coil. The current increased by 1/5 which is directly proportionate to the Voltage increase.

Exercise 5E, Case 5

The core is removed, what will be the voltage and current out of the secondary vs time?

I predict that if we remove the core, both the voltage and current will decrease. The answer explains that this is because the core serves to transfer the magnetic field and electrical power between the coils.

Exercise 5E, Case 6

Back to having a normal transformer with a core. If the primary coil has 100 turns of wire and there is 1200 V across it, design a transformer that will produce 120 V at the secondary.

To solve this one we can see that 1200V/120V=10; meaning we need to use a factor of 10 to produce the 120 V at the secondary. So we need a coil 1/10 the size of the 100 turn coil, we need a 10 turn coil.

My prediction was accurate.

Exercise 5E, Case 7

With the same primary coil as in Case 6, how could you modify or add to this transformer to provide 120 V to five different houses independently?

I predict that is we put five different 10 turn coils onto this core we could provide 120 V to five different locations. This prediction was correct because they would all share the same magnetic field.