Voltage Drop in a Series Circuit with 100V & 2 50 Ohm Resistors

[Drakkith]

Ok, let's say you have a 100v voltage supply and two 50 Ohm resistors in a series circuit. The total resistance is 100 Ohms, leading to a current of 1 amp.

Now my question is, what exactly is voltage drop? I know the voltage drop across each resistor is 50 volts. If we were to look at one of the resistors, would it act AS IF it were it's own little circuit with a 50v voltage supply? (As 50 volts divided by 50 ohms is 1 amp, just like the "parent" circuit)

 

[Ratch]

Drakkith,

Now my question is, what exactly is voltage drop?

You first have to know what voltage is. It the the energy density of the charge (volts=joules/coulomb). To send charges through a conduction path, the charge energy density has to be greater at one end than the other end. As the charges travel through the resistor conduction path, they encounter collisions with the ionic core of the material, and lose energy in the form of heat. Arriving at the other end, they have less energy per charge, and by definition have lost voltage. That is what a voltage drop is about--less energy density per charge because of energy loss due to thermal dissipation in the resistor.

If we were to look at one of the resistors, would it act AS IF it were it's own little circuit with a 50v voltage supply? (As 50 volts divided by 50 ohms is 1 amp, just like the "parent" circuit)

A resistor loses energy. It never supplies energy like a voltage source does.

Ratch

 

[Drakkith]

You first have to know what voltage is. It the the energy density of the charge (volts=joules/coulomb).

What's energy density?

A resistor loses energy. It never supplies energy like a voltage source does.

Ratch

Of course.

 

[Ratch]

Drakkith,

What's energy density?

You mean the energy density of the charge? Measured in volts (joules/coulomb)? It takes energy to gather some electrons scattered around free space into one unit volume. They all have a negative charge, and repel each other. It takes more energy to gather them into a smaller space than before, and it takes more energy to gether more electrons into the same space as before. So the energy it takes to clump them together divided by the number of electrons clumped together is the energy density of the charge (volts).

Ratch

 

[NascentOxygen]

I know the voltage drop across each resistor is 50 volts. If we were to look at one of the resistors, would it act AS IF it were it's own little circuit with a 50v voltage supply? (As 50 volts divided by 50 ohms is 1 amp, just like the "parent" circuit)

Yes. It really can't be anything else. How could a resistor know what else is going on around it? It can't. A volt is a volt, regardless of how it came to be.

 

[Drakkith]

How could a resistor know what else is going on around it? It can't. A volt is a volt, regardless of how it came to be.

I don't think I understand what you're getting at here.

Edit: By that I mean that I don't understand the part about the resistor not knowing what's going on around it. That seems...obvious? Perhaps I'm misunderstanding what you're saying.

 

[NascentOxygen]

I don't think I understand what you're getting at here.

Edit: By that I mean that I don't understand the part about the resistor not knowing what's going on around it. That seems...obvious? Perhaps I'm misunderstanding what you're saying.

Perhaps I misunderstood what you were asking.

 

[russ_watters]

Ok, let's say you have a 100v voltage supply and two 50 Ohm resistors in a series circuit. The total resistance is 100 Ohms, leading to a current of 1 amp.

Now my question is, what exactly is voltage drop? I know the voltage drop across each resistor is 50 volts. If we were to look at one of the resistors, would it act AS IF it were it's own little circuit with a 50v voltage supply? (As 50 volts divided by 50 ohms is 1 amp, just like the "parent" circuit)

Yes.

 

[Drakkith]

Thanks all!