1. ## DC Electrical Circuit

I need to find the equivalent resistance and the current at I

I'm not sure where to start, I need to use Thevenins theorem? And it's a Wheatstone bridge?

If I take out the diagonal 3ohm resister which represents the load (I think this is the first step?) I'm not sure what to do next?

2. Use Kirchhoff's laws to write down 3 loop equations for voltage drops, and 3 node equations for currents.

3. OK hears what I've attempted to do;

Is any of that correct so far?

If it looks like I don't know what I'm doing please point me in the right direction. I'm starting to dislike this Kirchhoff chap!

Thanks

4. Here's a way to cut down on the work a bit. Use the rules for combining resistors in series or in parallel to simplify the circuit. For a start, the two 3Ω resistors that I have labelled in blue in the attachment are equivalent to a single 6Ω resistor. That can be combined with the green 3Ω resistor, using the rule for combining resistors in parallel, to form a single resistor. That single resistor is in series with the yellow 2Ω resistor, so again you can reduce them to a single resistor.

You then have a simplified circuit which has two links going from c to d in the diagram: there is a current $\displaystyle I_2$ through the 1Ω resistor $\displaystyle R_3$, and a current I_3 through the equivalent resistor that you have formed from the blue-green-yellow resistors. Now you start to use Kirchoff's laws, but with only three variables $\displaystyle I_1,\,I_2,\,I_3$ instead of five.

5. Alex - I'm glad you're making a game attempt at this, but your equations are somewhat off.

First,
I1 != I2 + I3 + I4 + I5 ( where != means does not equal )

Is it clear that segments DA, AB and BC all have I1 flowing in them? If not, think of a bus leaving node D and going counterclockwise. There is no place for the current to change until it hits node C. That's why the current is the same in all those segments. Once you hit node C, you see that all the current going into node C must come out of there. That gives:

I1 = I2 + I3.

Likewise the node E equation is:

I3 = I4 + I5

If you ask what the node F equation is, it is:

I5 = I5 (I5 going in, I5 going out).

Can you write down what the node D equation is?

Your loop equations ( for voltage ) are better. However you need to correct what current is running in R2 and R4. Also, you must make sure that the sign of the voltage drop is consistent (is a positive drop with or against current?).

You can also use OpAlg's suggestion to simplify the circuit somewhat - combine resistors in series (R4 and R7), then in parallel (R5 and the effective (R4/R7)), etc. Still, you do need to get familiar with writing down general Kirchhoff equations for a general circuit.

6. Brilliant! I'm making sense of it all now :-) I may have cracked it, thanks very much!

7. If you can combine resisters 4, 5, 6 and 7, can you not then combine that equivalent resister with R3 useing the parallel rule then with R1 and R2 in series? Then just use ohms law?

8. Originally Posted by AlexTweed
If you can combine resisters 4, 5, 6 and 7, can you not then combine that equivalent resister with R3 useing the parallel rule then with R1 and R2 in series? Then just use ohms law?
That depends what you're looking for. I thought that the original problem was to find the current flowing through the bottom right-hand segment of the diagram (the current I that you subsequently called $\displaystyle I_3$, or actually $\displaystyle -I_3$ since it goes in the opposite direction). If you combine all the resistors into a single one then you can only find the current $\displaystyle I_1$, because that would be the only one left.

9. It looks like you've gotten the hang of it. Yes, you can continue to simplify the circuit in the way you described. However, as OpAlg points out, you've just simplified your problem out of existence! So back up one step and I think you're there. Good work!