when we talk about having a 'voltage drop' when current flows through a resistor.. how do we express that mathematically?

i guess i dont even understand the concept of if we have current flowing through these two resistors

http://img232.imageshack.us/img232/6057/123123p.jpg

there is a voltage drop from a to b then b to c right? how do we express that mathematically?

also, what would be the potential difference from c to the negative terminal of the voltage source? a short circuit so 0v?

thanks

 

Yeah, it took me awhile (all those decades ago) to get that strait. It's like this: The battery applies a voltage across the "circuit". In this case the circuit is composed of two resistors in series. The wires are also "resistors" which are also in series (with the resistors and the battery). So, you have a circuit composed of the wire that connects the + terminal of the battery to the 'a' side of the 2 ohm resistor, the 2 ohm resistor, the wire that connects the 'b' side of the 2 ohm resistor (stick with me, kid, there's a point to all this to the 'b' side of the 3 ohm resistor, the 3 ohm resistor, and finally the wire that connects the 'c' side of the 3 ohm resistor to the - terminal of the battery.

Now Kirchhoff's second law states that directed sum of the electrical potential differences around any closed circuit must be zero. So, if we go around the "loop" and add up the "potential differences" (also called the "voltage drops") it looks like this:

10V = 0V + 4V + 0V + 6V + 0V

The '0V' quantities are the wires. The wires drop (relatively) zero volts because their resistance is effectively zero compared to the resistance of the resistors. In actuality the resistance of the wires is NOT zero, more like, perhaps .00001 ohms [depends on what the wire is made of, how thick it is, and how long it is], but in most cases the resistance is going to be so low as to be negligible. THUS, once you get it that the wires drop (appreciably) no voltage, you can ignore the wires and just focus on the resistors:

10V = 4V + 6V or 10V = 2Amp(2ohm) + 2Amp(3ohm)

Another way to kind of get the picture is to imagine taking a length of NiChrome wire (a type of wire that has relatively high resistance often used to make wire wound resistors) and connecting it across the battery (such that one end of the wire is connected to the + terminal and the other end is connected to the - terminal. Now, imagine taking a volt meter and placing the neg probe at the - terminal of the battery, (making sure not to burn yourself, because that wire is probably going to get a bit hot) and then place the positive probe at the + terminal. The meter should read 10 volts. Now take the pos probe and start sliding it along the wire towards the - terminal. Notice how, as you move the probe, the voltage starts to decrease. When you are 3/4 of the way around, the voltage is going to be 3/4 * 10 or 7.5 volts. At half way around the reading is going to be 5 volts, etc.

It's the same sort of thing with the little circuit you drew (with the 2 ohm and the 3 ohm resistors). The only difference is that the resistance isn't uniform around the circuit. As you move the pos probe around the your little circuit, you will see 10 volts until you reach the ‘a’ side of the first resistor (the 2 ohm resistor), then when you get to the other side of the 2 ohm resistor the meter will read 6 volts--and will keep reading 6 volts until you reach the ‘b’ side of the 3 ohm resistor. Now, let’s say you somehow managed to expose the resistive material in that resistor (e.g. you cut it open). With the resistive material exposed, when you run the probe over, say, the 2 ohm resistor, you will see 10 volts at the 'a' side and as you run the probe along the 2 ohm resistor the meter reading will gradually lower until it reaches 6 volts, then it will stay at 6 volts as you run the probe over the wire that connects the 2 ohm resistor to the 3 ohm resistor, then as you run the probe along the 3 ohm resistor the voltage will gradually fall until it reaches zero volts. It will remain 0 volts all the way to the - terminal on the battery.

Now, just to either confuse you, or really make this whole thing crystal clear. If you place the negative probe on the + terminal on the battery and place the positive probe on the - terminal, you will see -10 volts (that's negative 10 volts). Then, as you run the probe down the wire to the ‘c’ side of the 3 ohm resistor (remember, you're going the opposite direction) you will continue to see -10 volts until you reach the 3 ohm resistor. As you run the probe down the exposed 3 ohm resistor you will see the meter reading gradually fall until it reaches -4 volts at the other side of the 3 ohm resistor, and well hopefully you get the idea.

In other words, the polarity is COMPLETELY ARBITRARY. It doesn't matter which way you go around the loop. Just pick a common (or ground) and then all your measurements will be relative to that point.

So, just remember that the wires are resistors, too, but you usually ignore them because their resistance is so low that the voltage they drop (and they do drop voltage) is usually negligible. There ARE exceptions, such as the lines that connect a high current power supply to a circuit. Often, a second pair of wires will be run called "sense lines" that allow the power supply to compensate for the voltage dropped in those lines, but that another lecture

The BEST way to get a sense of this is to get an actual battery, a couple of resistors (probably better to use, say 2k and 3k so you’re less likely to see smoke) and a meter and probe the heck out of it, until you get used to how it works.

 

Thanks Puppy... that clears it up a bit... but I'm still confused with mathematically analyzing it. It apparently has to do with KVL though.

How did you get the 4V and 6V?

 

The total current throughout the circuit is calculated quite easily.

Add the resistors together, 2+3=5
So the total resistance is 5Ω

By Ohm's Law V=I*R
So V=10, R=5

V/R=I , 10V/5Ω=2Amps

2 Amps is the current throughout the circuit.

I*R=V, so just multiply the current over each resistor to get
the 4V drop over the 2Ω resistor and the 6V over the 3Ω resistor.

Also the application of voltage divison lends a hand.
This is basically the same math as shown above albeit reduced down to one equation

V=[Vs(R1)]/R1+R2 , where Vs is the source voltage and R1 and R2 are resistors in series.
Try both out and see if you understand the use of voltage divison.

 

Think I got it, thanks.