# Simple Circuit Diagram (Ohm's Law Application)

1. Aug 12, 2009

### exitwound

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

Consider the circuit shown below. (a) Determine the current reading of the ammeter.
(b) Determine the voltage reading of the voltmeter. (c) Determine the power dissipated by the 500 W resistor.

What would be the first steps in attempting this?

There are differing resistances on the three branches of the circuit. Therefore, I have different amounts of current on each branch. As a result, I can't figure out the voltage drop across the first resistor.

I don't know how to proceed.

2. Aug 12, 2009

### rock.freak667

You can use Kirchoff's laws OR find the combined resistance, then find the total current and go from there.

3. Aug 12, 2009

### exitwound

I don't know how to find the total resistance in this situation.

I know Resistance in series = R1 + R2 + R3 + ... Rn
resitance in parallel is 1/R = 1/R1 + 1/R2 + 1/R3 + ... 1/Rn

I don't know where to do the equivalent resistance substitutions in this diagram.

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

### bm0p700f

A parrallel circuit is like series circuits joined together. How many series loops do have in that circuit?

What is the total p.d in each branch of this parallel circuit.

What is the current going thorough the 330ohm resistor?

What is the current going through the 250 ohm resistor (be careful here)?

You don't need to combine resistors really for this question.

What equation must you use to find the above?

The the final bit is easy.

5. Aug 12, 2009

### exitwound

Ohm's Law. V=IR.

I don't know how to start the problem. Having a voltmeter on a branch with no resistance is bugging me. Having an ampmeter measuring the current on one branch is something else I can't figure out.

You say I don't have to combine resistances, but how can I figure out the current if I don't?

6. Aug 12, 2009

### Staff: Mentor

A voltmeter is a very high input impedance measuring device. So you can treat the voltmeter as an (open or short?) circuit....

A current meter is a very low input impedance measuring device, so you treat it as a ? circuit...

Once you make those substitutions, you can combine the parallel resistances on the right side, figure out the total current, then the voltage across the parallel resistances, then the current through each leg. Give all that a try, and post your work here.

7. Aug 12, 2009

### exitwound

Well, you can combine the resistances on the right side to 142.24 Ohms, then add it to the serial connection with the 500 Ohms for a total of 642.24 Ohms of resistance in the circuit.

Using that, we can find that the current flowing through the circuit as a whole is 9.34e-3 Amps.

Then we find the voltage drop across the first resistor. (9.34e-3)(500)=4.67V.

Use the 1.33 Voltage on the 250Ohm resistor I=V/R=532mA.

The analogy that seemed to help me the most was from an online friend. And it's funny, I'll probably try to put all my faith into Vikings from now on:

8. Aug 12, 2009

### Staff: Mentor

How in the world is that quote relevant here?

9. Aug 12, 2009

### exitwound

It's how he described current, voltage, and resistance. He doesn't subscribe to the water analogy either because he said it's more problematic than helpful. I can't use the water analogy either. It just doesn't make sense in my head. But when he described it as Vikings, it was a little easier to understand what he meant. Why? I have NO idea. but it worked for now.

10. Aug 12, 2009

### Paulo Serrano

Am I wrong to say that the voltage stays constant throughout the entire circuit?

11. Aug 12, 2009

### vk6kro

Am I wrong to say that the voltage stays constant throughout the entire circuit?
Yes, that would be wrong.

Maybe the meters are freaking you out.

You have a resistor in series with two resistors in parallel.
You know how to work out the equivalent resistance of two resistors in parallel.

So then you have two resistors in series.
Can you work out the voltages across them?

So, what would the voltmeter read?

All of the resistors now have a known voltage across them.
What current would flow through the 250 ohm one?

What power would be dissipated in the 500 ohm resistor?

12. Aug 13, 2009

### Paulo Serrano

I know very little about circuits, but isn't there something that stays constant throughout the entire thing? Since you say it's not the voltage, could it possibly be the current?

13. Aug 13, 2009

### vk6kro

Paulo....
Yes, in a series circuit, the current in all components is the same.

In this case, however, there were parallel resistors that were not equal, so the currents in these were not the same.

This is an excellent problem if you would like to have a go at solving it. Why not try if you have an interest in circuits? Send me personal mail if you have need help with it.

14. Jun 28, 2010

### justing007

is it 4.5 volts!! across 500.. 250 and 330 is 1.5... current in 500 is .009 - - 250 is .006 and 330 is .004..

series

Et=E1+E2+E3.....
It=I1=I2=I3.......

Parallel

Et=E1=E2=E3......
It=I1+I2+I3.......

just a guess.... Use kirchoff's law!!

15. Jun 28, 2010

### vk6kro

You would work out the combined resistance of the 330 ohm and 250 ohm resistors. These are in parallel. Call this R. (but work it out exactly)

Now add this resistor instead of the 330 and 250 ohms. So, you get 500 + R.

Now you can work out the total current. It is = E / R or (6 volts / 500 + R)

Then you can work out the voltage across R. This is the answer to part B.

This is the same voltage as across the 250 ohm resistor. So, you can work out the current in this resistor. This is the answer to part A.

This thread is nearly a year old and the answers were actually given in earlier posts, as well as explanations of how to solve it.