# Resistors in Series (Ohm's Law)

1. Aug 12, 2009

### exitwound

1. The problem statement, all variables and given/known data

Three resistors are connected to a 9.00 V battery: R1 = 100 W, R2 = 200 W, R3 = 300 W. Determine the voltage, current, and power for each resistor for a series connection.

2. Relevant equations

V=IR

3. The attempt at a solution

$$V_{total}=I_{total}+R_{total}$$
$$\frac{9}{600}=I_{total}=.015A$$

However, from this point I don't know how to separate the individual resistors using Ohm's Law. I have for each resistor:

$$V_1=(I_1)(R_1)$$
$$V_2=(I_2)(R_2)$$
$$V_3=(I_3)(R_3)$$

I also know that

$$I_{total}=I_1+I_2+I_3$$

I just don't have enough equations.

Is it feasible to say that since 1/6th of the total resistance is being used on the first resistor that 5/6ths of the voltage is being 'used' there? And that over the 200 Ohm resistor, 1/3rd of the total voltage is being 'used'? and over the 300 Ohm resister, 1/2 the voltage is being 'used'? Is it as simple as this or is it better to set up a series of equations as above? The math works, but I don't understand the reasoning.

I think I'm missing a simple concept here that'll plug the hole in my head regarding the circuit.

2. Aug 12, 2009

### Cyosis

The reason why you are having difficulties is that the formula for $I_{total}$ is wrong. Remember current does not get used up. What does this imply for I1, I2 and I3? If you don't know replace the current with water in your mind and the resistors with pipes of different diameter. If you throw 10 liter/s in on top how much is going to come out at the bottom? How much water has to flow through each pipe?

Last edited: Aug 12, 2009
3. Aug 12, 2009

### exitwound

I don't know, that's the problem. Obviously the amps going into the circuit have to come out of the circuit. But if each resistor is using up a different amount, how do I compute one individually?

Would I_total above be right if I had a parallel setup of resistors? Wouldn't the water going through pipe #1, pipe #2, and pipe #3 all have to add up to the amount going in and coming out?

4. Aug 12, 2009

### mikelepore

You're asking whether there is a proportional dividing-up of the 9 volts among the three resistors, such as whether the voltage across R1 will be

(9 volts) (R1 / [R1 + R2 + R3])

etc.

To find out whether that's logical, you can try that and then verify it by performing these two checks:

(1) See whether the three resistor voltages add up to 9 V

(2) Look at the current values (the ratio V/R) for each of the three resistors. How do they all compare? Does that result make sense?

5. Aug 12, 2009

### exitwound

I'm obviously missing something. I don't understand what's going on so plugging numbers into equations isn't making it easier to grasp. I need to understand how it works first, then I can worry about putting numbers in. But i'm obviously in lack of something crucial to make it make sense in my head.

6. Aug 12, 2009

### bm0p700f

You have to first work out the current. Cynosis description is very accurate model for current. I will try putting it another way. The formula you gave "the sum of the individual current = total current works a junction in a circuit2 i.e I(tot) = I1 +I2 +I3.

You have a series circuit. Do you have a junction in you series circuit? What do you know about the current in a series circuit?

Some people see the water model clearly other like this model. So here goes.

If you have chain driven by a large chain ring. This is your battery. The chain runs over three different size sprockets. Each sprocket (gear) is like you resistor. The moving chain is your current. You turn the large chain wheel. Do you have the same amount of chain leaving the large chain wheel as you have returning to it. What do you know about the speed of the chain in this circuit? Does it change....?

Apply this model or Cynosis to your series circuit. Then tell us what the current through the resistors and how you worked it out. The rest is easy really.

7. Aug 12, 2009

### tms

So far so good.

Not so good. Think more carefully. If the all the current that goes in one end comes out the other, how can resistors in series "use up" different amounts of current? Obviously, they can't. It is the idea of resistors "using up" current that is causing your trouble. I think you are mixing up current and voltage.

8. Aug 12, 2009

### exitwound

If we think of it as a chain,

moving Chain = current
sprocket size = resistance
sprocket rotation = voltage

the chain itself has to move at a certain velocity over the different sized sprockets. The rotation of the sprockets is the only thing that can change. I'm guessing that the voltage is the rotational speed of the gears in this analogy? If that's the case, the only thing that can change in the entire scenario is the voltage.

If that's the case, after the first resistor, the potential 'left' from the 9V original source has been diminished by the first resistor.

If I use the equivalent resistance of 600 Ohms, and a 6V source, V=IR, I=V/R, I=.015 Amps. This is the current that flows in this scenario.

The only thing that changes across a resistor is the potential then? If I picture it as a chain, I can see how it works.

But if it's water:

flowing water: current
resistors: size of pipe
voltage: amount of water?

I can't picture the water analogy at all.

9. Aug 12, 2009

### Cyosis

The current is the water current in that analogy. The first thing that you have to understand and never forget is that current does not get used up nor does water. Imaging you pouring in 10 liters of water and zero coming out of the bottom you would be baffled would you not? Every single pipe will see the same amount of water going through it and this is the same for the current.

10. Aug 12, 2009

### vk6kro

You have one equation missing.
The total resistance of resistors in series is the sum of the resistors.
so, RTotal = R1 + R2 + R3

So, add them up and work out the total resistance. You already did this.

Now, what current is flowing from the battery? You already did this. Does it flow through all the resistors? (Yes, that is one property of a series circuit)
So you don't need this:
Itotal =I1 + I2 + I3. There is only one current.

So, you know the resistance and you know the current in each resistor. How do you work out the voltage?
Then how do you work out the power?

11. Aug 12, 2009

### exitwound

If you know the current and the resistance, then the voltage can be found by the relationship V=IR, and the power can be found by using V^2/R.

When talking about electricity, we often use 'power' in an everyday fashion. But what exactly is POWER when talking about a circuit such as the above?

12. Aug 12, 2009

### vk6kro

Yes, you have those equations OK.

Power means the same thing as in real life.
In a resistor, it is a heating effect. It might be small in this case, but it is just the voltage across a resistor times the current through it.

13. Aug 12, 2009

### mikelepore

For an energy source like the battery, power is the rate at which it supplies energy to the circuit.

For a resistor, power is the rate at which it dissipates energy away from the circuit, and that energy becomes heat.

One watt of power means a rate of one joule of energy per second (1 W = 1 J/s).

The expression for power P = VI can be used for both a battery and a resistor.

Additional expressions for power P = I^2 R and P = V^2 / R can be used only for a resistor. All three formulas for resistor power will give the same answer.

14. Aug 12, 2009

### mikelepore

The water analogy is, if you have three pipes or garden hoses connected end-to-end, the water would go into the first pipe, then come out of it and go into the second, then come out of it and go into the third. That's a way of picturing why three resistors in series have the same current going through them. If you had pictured that you wouldn't have written I total = I1 + I2 + I3. There is only one current in this circuit. There is no total.

15. Aug 13, 2009

### vk6kro

I noticed this in your original post:
Is it feasible to say that since 1/6th of the total resistance is being used on the first resistor that 5/6ths of the voltage is being 'used' there? And that over the 200 Ohm resistor, 1/3rd of the total voltage is being 'used'? and over the 300 Ohm resister, 1/2 the voltage is being 'used'? Is it as simple as this

Can you now see that this is wrong?
All resistors have the same current flowing in them so the voltage across them is (I * R) and directly proportional to the resistance.

Assuming a current of 15mA (0.015 amps),
the 100 ohm has 100 * 0.015 or 1.5 volts across it.
the 200 ohm has 200 * 0.015 or 3 volts across it.
the 300 ohm has 300 * 0.015 or 4.5 volts across it.
These voltages, 1.5, 3 and 4.5 add up to 9 volts as you would expect.