• krtica
Yes. The three resistors in parallel share nodes (on the ends of each resistor), thus they share a common potential. You need to find the "voltage drop" across the 3 Ohm resistor first though. But again, recheck your "(126/27)" figure. I think it's just a little off, if I did my math correctly.The three resistors in parallel share nodes (on the ends of each resistor), thus they share a common potential. You need to find the "voltage drop" across the 3 Ohm resistor first though. But again, recheck your "(126/27)" figure. I think it's just a little off, if I did my math correctly.The voltage

#### krtica

http://www.webassign.net/tipler/26-50.gif

I presume the 4 ohm resistor is parallel with the diagonal 2 ohm. It's equivalence is parallel to the vertical 2 ohm. Then, the result is in series with the 3 ohm.

I then substituted values in V=IR and worked backwards to find the current and remaining voltage. Please explain the process.

What exactly are you trying to find? If it's current, then you're doing it the right way. What's "remaining voltage"?

ideasrule said:
What exactly are you trying to find? If it's current, then you're doing it the right way. What's "remaining voltage"?

Yes. By "remaining voltage" I mean the potential difference between each resistor beginning with the first.

In that case, you're doing it the right way: find the current through its resistor, then use V=IR.

ideasrule said:
In that case, you're doing it the right way: find the current through its resistor, then use V=IR.

Thank you so much.

I'm not sure exactly what I am doing wrong.

The current for the 3 ohm resistor is 42/27, which came up correct.

I assumed that since the 3 ohm resistor and the 2 ohm resistor share the same branch, they have the same current, which isn't correct.

For resistors in parallel, would I calculate the potential difference between the original 6 volts and the potential of the 3 ohm (126/27), then take the value and divide by each resistor in a parallel combination?

krtica said:
I assumed that since the 3 ohm resistor and the 2 ohm resistor share the same branch, they have the same current, which isn't correct.

It's not correct because some of the current flows that goes through the 3 ohm resistor into the other 2 ohm resistor, and some flows into the 4 ohm resistor.

For resistors in parallel, would I calculate the potential difference between the original 6 volts and the potential of the 3 ohm (126/27), then take the value and divide by each resistor in a parallel combination?

No, that value IS the voltage drop across each resistor.

krtica said:
I'm not sure exactly what I am doing wrong.

The current for the 3 ohm resistor is 42/27, which came up correct.

You might want to recheck that. Your answer might be close enough for your answer to be accepted within some reasonable error, but the fraction that I came up with is not 42/27. The resulting decimal equivalent that my fraction produces is quite close, but not identical.

This small difference might end up being part of the problem with why the rest of the answers aren't working.

I assumed that since the 3 ohm resistor and the 2 ohm resistor share the same branch, they have the same current, which isn't correct.

I'm not sure I follow you here. After the current passes through the 3 Ohm resistor, it splits into three parallel branches.

For resistors in parallel, would I calculate the potential difference between the original 6 volts and the potential of the 3 ohm (126/27), then take the value and divide by each resistor in a parallel combination?

Yes. The three resistors in parallel share nodes (on the ends of each resistor), thus they share a common potential. You need to find the "voltage drop" across the 3 Ohm resistor first though. But again, recheck your "(126/27)" figure. I think it's just a little off, if I did my math correctly.

collinsmark said:
You might want to recheck that. Your answer might be close enough for your answer to be accepted within some reasonable error, but the fraction that I came up with is not 42/27. The resulting decimal equivalent that my fraction produces is quite close, but not identical.

This small difference might end up being part of the problem with why the rest of the answers aren't working.

In regard to how I found the fraction, I'll show each step.

The 4 ohm and diagonal 2 ohm are in parallel, and I found their equivalence to be 4/3 ohm. This equivalence is parallel with the vertical 2 ohm, which then is 4/5 ohm. The 3 ohm and the last equivalence are in a series and equal to 19/5 ohm.

I=V/R; I=6/(19/5); I= 30/19? 42/27 may have been a miscalculation.

collinsmark said:
I'm not sure I follow you here. After the current passes through the 3 Ohm resistor, it splits into three parallel branches.

Could I find the vertical 2 ohm resistor's current by V=(I_1*R_1)+(I_2*R_2), where the subscript 1 denotes the 3 ohm resistor and the subscript 2 denotes the vertical 2 ohm resistor?

collinsmark said:
Yes. The three resistors in parallel share nodes (on the ends of each resistor), thus they share a common potential. You need to find the "voltage drop" across the 3 Ohm resistor first though. But again, recheck your "(126/27)" figure. I think it's just a little off, if I did my math correctly.

Would the voltage drop be 6V-(90/19)V? Then divide this quantity by each resistor in parallel (the diagonal 2 ohm with the 4 ohm)? Or would I calculate the voltage drop again after finding the current for the diagonal 2 ohm resistor in order to find the 4 ohm resistor's current?

This means a lot. Thank you for your help.

ideasrule said:
It's not correct because some of the current flows that goes through the 3 ohm resistor into the other 2 ohm resistor, and some flows into the 4 ohm resistor.

Could I find the current in the 2 ohm resistor by V=(R*I)+(r*i), where R and I are the current and resistance for 3?

krtica said:
In regard to how I found the fraction, I'll show each step.

The 4 ohm and diagonal 2 ohm are in parallel, and I found their equivalence to be 4/3 ohm. This equivalence is parallel with the vertical 2 ohm, which then is 4/5 ohm. The 3 ohm and the last equivalence are in a series and equal to 19/5 ohm.

I=V/R; I=6/(19/5); I= 30/19?

Now you're cookin'.

Could I find the vertical 2 ohm resistor's current by V=(I_1*R_1)+(I_2*R_2), where the subscript 1 denotes the 3 ohm resistor and the subscript 2 denotes the vertical 2 ohm resistor?

Hmm. If I'm following you correctly, then technically yes, this way will lead to the correct answer, as long as you recognize that V here is equal to 6 Volts. Then solve your equation for I_2.

But that method is sort of doing a few things in one big step. There's nothing necessarily wrong with it, but it might be easier to break up to the problem a little as described below.

Would the voltage drop be 6V-(90/19)V? Then divide this quantity by each resistor in parallel (the diagonal 2 ohm with the 4 ohm)?

There you go. Doing it this way is a little bit simpler than the other way discussed above, but it should give you the same answer (I like this way better).

You now know the voltage across the rest of the resistors (the two 2 Ohm, and the single 4 Ohm). Once you know the voltage across a resistor, simply divide by its resistance to find the current through that resistor.

But just to double check, you might want to do it this way, and the way discussed above to make sure you get the same answer. (mathematically, they are really the same thing)

Or would I calculate the voltage drop again after finding the current for the diagonal 2 ohm resistor in order to find the 4 ohm resistor's current?

There's no need to recalculate the voltage drop. All three resistors have the same voltage drop.

Don't worry about that diagonal way in which the resistor was drawn. It may have been drawn that way just to throw you. If it helps, re-draw the circuit so that each resistor is either horizontal or vertical (but not diagonal). After doing so it is easy to see both the 2 Ohm resistors and the 4 Ohm resistor all have the same potential (and all three resistors are in parallel).

Thank you! You're explanations could not be simpler.

Would the power delivered by the battery be the summation of each current multiplied by the voltage?

Yes.

krtica said:
Would the power delivered by the battery be the summation of each current multiplied by the voltage?

Yes, conservation of energy applies here. The total power delivered by the battery must equal the sum of the powers dissipated by each resistor. One way to do that is to multiply the current through a given resistor by the voltage drop across that particular resistor; then repeat and sum for all resistors in the circuit (using their respective currents and voltage drops).

I understand. Thank you, again.