# Homework Help: Electrical Circuit

1. Jan 11, 2006

### Gamma

I had a complicated electrical circuit which I was able to reduce to the following (attached). How do I go about finding the resultant resistance of this circuit? Is there a short way of doing this?

Thanks,

Gamma

#### Attached Files:

• ###### Circuit.JPG
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2. Jan 11, 2006

### Gamma

Using Kerchoff's law would be complecated. I tried that. Did not proceed with the calculation. I am sure there is an easy way. If some body can show me how I would proceed that would be great.

Thanks.

3. Jan 11, 2006

### Gamma

I reduced it by allowing a known current (1 amp) to pass through the complecated circuit (not the R and 2R which is in series with the battory).

Finally I am getting a number which is 3R + 16/50 R.

4. Jan 12, 2006

### robphy

For this problem, you should be able to replace a pair of resistors in parallel or series by its equivalent resistance. Continue this process until you end up with one equivalent resistance.

5. Jan 12, 2006

### Gamma

The circuit in the lower right hand coner seems that it can not be reduced using parallel/series method. If you look at a junction, For example the junction of 4R/3 and R/2 ( in the far right), the current does not simply devide between these two resistors. Rather the current at this juction sees 4R/3 in one branch and some other effective resistance in the other branch. So can I take 4R/3 and R/2 as parallel?

6. Jan 12, 2006

### andrewchang

yes, but there's other resistors that should also be included.

7. Jan 12, 2006

### BobG

A lot of times it helps to redraw this in a way that makes it easier to see each branch before trying to calculate anything.

From the positive terminal, you pass through a 2R resistor. Then you come to a junction that branches off. One of those branches is a straight line that goes to another junction. If you redraw this circuit, the straight line does you no good at all. What you really have is a junction with three different branches. Of the three branches, one has a resistor in series with two parallel resistors. The other two branches have one resistor each.

Once you reduce the more complicated branch down to one equivalent resistance, your problem gets a lot simpler. You'll have three resistors in parallel with each other.

And, of course, you finally go through one more resistor in series.

8. Jan 12, 2006

### Gamma

Thanks every one. Redrawing helped simplify the problem a lot.

Finally I am getting a number which is 3R + 4/25 R which is same as what I got before, but it took less time.

regards,

gamma