Voltage and Current Clarification

[.:Endeavour:.]

I need some clarification with voltage and current and how you calculate the voltage and the current. Also when you are given the voltage and the current, how to find the charge and the time for the current; and the work and the charge for voltage.

This is the formulas that I have for voltage and how to get the other elements, like W and Q, in order to calculate the voltage.

$$V = \frac{W}{Q}$$

To find W or the work you do this formula, I think:

$$W = Fd$$

To find Q or the charge you do this formula:

$$Q = \frac{number of electrons}{6.25 * 10^1^8}$$

To find the current of a wire or a conductor you use this formula:

$$1A = \frac{1C}{1s}$$

$$1C = 1A * 1s$$


I'm not sure if to find the W for the voltage you use the formula above or not. Where can I find a chart or graph comparing other types of currents and voltage for example the current and voltage of a TV set to a blender. Thank you for your time.

 

[Dale]

Where can I find a chart or graph comparing other types of currents and voltage for example the current and voltage of a TV set to a blender.

Almost every appliance that you plug in will have a little "UL" sticker on it that will tell you the voltage and current that the appliance is designed to operate at. Sometimes it is listed as voltage and power, in which case simply divide the power by the voltage to get the current.

 

[.:Endeavour:.]

Almost every appliance that you plug in will have a little "UL" sticker on it that will tell you the voltage and current that the appliance is designed to operate at. Sometimes it is listed as voltage and power, in which case simply divide the power by the voltage to get the current.

So, like my lamp that has a max of 60 Watts. To find the Work you do Work = Power * time (in hours). It will look like this:

60W * 3600s = 216,000 Joules or 216 kJ

The amount of work is 216 kJ that then gives you 60 Watts of power in one hour. The only thing that I'm confused is finding the work and the charge that equals the voltage for example 10V.

 

[Dale]

So, like my lamp that has a max of 60 Watts. To find the Work you do Work = Power * time (in hours). It will look like this:

60W * 3600s = 216,000 Joules or 216 kJ

The amount of work is 216 kJ that then gives you 60 Watts of power in one hour. The only thing that I'm confused is finding the work and the charge that equals the voltage for example 10V.

OK, in the US the voltage is 120 V. So the total charge over the hour is 216 kJ/120 V = 1800 C.

 

[.:Endeavour:.]

Ok, let's say that you have a charge of 10μC and you wanted to find the work so then you can find the voltage. Do you do the W=Fd formula to find the work?

 

[Dale]

Usually you go the other way, you have a defined voltage and then you calculate the work done. But sure, you could do that if you had a well defined F and d.

 

[.:Endeavour:.]

Oh, ok. To get the force, you use Coulomb's Law, but I'm not sure if the distance is the length of the wire or how far you are going to do the work. That's what has me puzzled. The formula that I have for Coulomb's Law is:

$$F = \frac{kQ_1Q_2}{d^2}$$

To find the work, you use the force that you get from Coulomb's Law and the distance that you to find the force as well to then find the work?

 

[Dale]

Oh, ok. To get the force, you use Coulomb's Law, but I'm not sure if the distance is the length of the wire or how far you are going to do the work. That's what has me puzzled. The formula that I have for Coulomb's Law is:

$$F = \frac{kQ_1Q_2}{d^2}$$

To find the work, you use the force that you get from Coulomb's Law and the distance that you to find the force as well to then find the work?

You would not usually use Coulomb's law, it is only for electrostatics and is not generally applicable. It is usually used to determine the force between two stationary point charges, not the force on an individual charge carrier in a current. The force in general is the charge times the E-field, but the E-field is just the gradient of the voltage so if you know the E-field then you know the voltage anyway.

I think you are getting confused because you are trying to go backwards. You are trying to take a known charge, do some known work on it, and determine the voltage. You will usually go the other way, take a known charge, put it through a known voltage, and determine the work done.

 

[.:Endeavour:.]

You would not usually use Coulomb's law, it is only for electrostatics and is not generally applicable. It is usually used to determine the force between two stationary point charges, not the force on an individual charge carrier in a current. The force in general is the charge times the E-field, but the E-field is just the gradient of the voltage so if you know the E-field then you know the voltage anyway.

I think you are getting confused because you are trying to go backwards. You are trying to take a known charge, do some known work on it, and determine the voltage. You will usually go the other way, take a known charge, put it through a known voltage, and determine the work done.

That's what my teacher said about Coulomb's Law and how its expressed. This is my other approach to find the charge. Let's say you have a current of 4amps. Time is normally expressed in hours for electronic equipment (mostly) so you just multiply the current with the time to get the charge like this:

Q = I * t
Q = 4amps * 3500s
Q = 14,400C

I've been looking at electrical devices and have found that the voltage is higher than the current, and I want to know does voltage determin the current or vice versa. I've looked over the fomula for voltage and current and noticed that they both are determined by Q. So if you want to find the voltage from a known current like 4 apms and know the time interval, can you use the same charge to find the voltage knowing the work?

 

[JaWiB]

If you know the work done, the current, and the time interval, couldn't you just use P=IV=W/t, thus V=W/It?

 

[Dale]

I've been looking at electrical devices and have found that the voltage is higher than the current, and I want to know does voltage determin the current or vice versa.

Neither determines the other. They are independent quantities. Together they determine the power, P=IV. Or rather, any two determines the other. Usually for appliances voltage is fixed at 120V and the power requirements are fixed by what the appliance needs to do and so that determines the current you need.

I've looked over the fomula for voltage and current and noticed that they both are determined by Q. So if you want to find the voltage from a known current like 4 apms and know the time interval, can you use the same charge to find the voltage knowing the work?

Yes, it is the same Q. So again, P=IV=(Q/t)(W/Q)=W/t as JaWiB mentioned also.

 

[.:Endeavour:.]

Ok, thank you for the clarification, both of you. This is my last question. When you look at the voltage of an electrical device, is it the energy needed to run the electrical device when having the proper current? This what I have in my book: " One volt is the potential difference between two points when one joule of energy is used to move one coulomb of charge from one point to the other."

 

[Dale]

When you look at the voltage of an electrical device, is it the energy needed to run the electrical device when having the proper current? This what I have in my book: " One volt is the potential difference between two points when one joule of energy is used to move one coulomb of charge from one point to the other."

Voltage is not energy. It is energy per unit charge or power per unit current.

The voltage of an electrical device is actually a design decision. For example, consider my wife's hair dryer. It has a specific purpose, it needs to heat up air and blow it out the nozzle. This takes a certain amount of power to do that job, specifically 1.8kW. Now, the manufacturers of the hair dryer need to design their device to consume 1.8kW of electricity. Since in the US the wall sockets provide 120V they base their design around that and make it draw 15A of current at 120V since 15A 120V = 1.8kW.

If I took that same dryer to Kuwait and plugged it in then, since Kuwait uses 240V it will draw 30A for a total power of 30A 240V = 7.2kW. The dryer will not be able to withstand that much power so it will burn out very quickly. The reason that it will burn out is because it is not designed to operate correctly at 240V. A properly designed hair dryer for Kuwait will draw only 7.5A of current at 240V in order to get the right power of 7.5A 240V = 1.8kW.

 

[.:Endeavour:.]

Voltage is not energy. It is energy per unit charge or power per unit current.

The voltage of an electrical device is actually a design decision. For example, consider my wife's hair dryer. It has a specific purpose, it needs to heat up air and blow it out the nozzle. This takes a certain amount of power to do that job, specifically 1.8kW. Now, the manufacturers of the hair dryer need to design their device to consume 1.8kW of electricity. Since in the US the wall sockets provide 120V they base their design around that and make it draw 15A of current at 120V since 15A 120V = 1.8kW.

If I took that same dryer to Kuwait and plugged it in then, since Kuwait uses 240V it will draw 30A for a total power of 30A 240V = 7.2kW. The dryer will not be able to withstand that much power so it will burn out very quickly. The reason that it will burn out is because it is not designed to operate correctly at 240V. A properly designed hair dryer for Kuwait will draw only 7.5A of current at 240V in order to get the right power of 7.5A 240V = 1.8kW.

Ok, now I get what your trying to say for voltage. So, like my lamp that needs to run at 0.06kW and the standard voltage is 120V. The current that will be 0.5amps to meet the 0.06kW need to run the lamp. One thing that I don't get on your explenation, if the hair dryer uses 15 amps to get the 1.8kW needed to run, why does it then run at 30 amps in Kuwait and not at 15 amps?

 

[Dale]

One thing that I don't get on your explenation, if the hair dryer uses 15 amps to get the 1.8kW needed to run, why does it then run at 30 amps in Kuwait and not at 15 amps?

Because of "[URL law[/URL]. If you haven't reached that yet in your class then don't worry, you will very soon. The resistance of the dryer is R=V/I or 120V/15A=8Ohms. That is how the designers make it draw 15A at 120V, by setting its resistance to 8Ohms.

 

[.:Endeavour:.]

Because of "[URL law[/URL]. If you haven't reached that yet in your class then don't worry, you will very soon. The resistance of the dryer is R=V/I or 120V/15A=8Ohms. That is how the designers make it draw 15A at 120V, by setting its resistance to 8Ohms.

That's what started to think about the resistance that the hair dryer has. One way to get the power back to 1.8kW is to get the current to 7.5 amps. The resistance then should be at 32 $$\Omega$$ because $$\frac{V}{I} = R$$. Because consumer products you can't actually change the resistance, the resistance will be at 8 $$\Omega$$.