# Someone help with simple circuit problem?

1. Mar 15, 2004

### ViEternal83

A series circuit with
R1=5 Ohms
R2=9 Ohms
V=10V

What is voltage drop across R2?

I got 11.11 Watts.

If someone can tell me if I'm wrong or not, it would be much appreciated.

Is this some sort of super smart people's forum where this kind of question is laughed at?

2. Mar 15, 2004

### Cliff_J

You shouldn't get watts for a voltage drop - watts is power. So your answer is automatically wrong because its incorrect units, and I don't get where you got that number but its wrong.

Few ways to solve this problem, here's one that should be easy to visualize. You know in a series circuit that the current will be identical across the components, so lets find that current and work with that using ohm's law.

Add the series resistance to get the total resistance. You know the voltage, find the current across this total resistance. Now you know the current you can find each voltage drop and both should add to equal your starting voltage.

If you really are looking for power dissapated by R2, its 4.59 watts. I'm not going to simply tell you the voltage drop, show your work and I'm sure someone will help you to learn how to do this yourself, very easy once you get the idea of using ohm's law and this is VERY valuable knowledge.

voltage = current * resistance

Cliff

3. Mar 15, 2004

### ViEternal83

Oh, haha.
I had no idea what I was doing apparently.
But thanks for telling me I was wrong and the information.
I'll try again.

4. Mar 15, 2004

### ViEternal83

Req = R1+R2 = 5 + 9 = 14 Ohms
I = V/Req = 10/14 = 5/7

Voltage drop across R2 = V = I*R2 = 5/7 * 9 = 6.43 Volts
Is this correct?

Thanks for the help.
What a nice forum.

Last edited: Mar 15, 2004
5. Mar 15, 2004

### Cliff_J

Yup, you got it now. Simple algerbra and you can just about figure it all out from ohm's law until you get to AC and non-linear behavior, but that's another story for a different day! :)

Cliff