# Help with circuits - resistors in parallel and in series

1. Feb 2, 2014

### chococat

1. The problem statement, all variables and given/known data
http://imgur.com/iTHbpTv
This picture contains the circuit drawn out, and all the provided numbers. It also includes some of my work.

2. Relevant equations
I=V/R
P=I*R

3. The attempt at a solution
So I attempted to reduce it down to one resistor & one battery (26V, 13 ohms).
I got that the current was 2Amps. I am having difficulty with resistors in parallel.
Honesty, I would appreciate more on how to tackle these types of questions that the actual answers (this is practice for an exam).
I feel really stupid because I don't know what I'm doing once there are resistors in parallel and in series added to the mix.

2. Feb 3, 2014

### Simon Bridge

Welcome to PF;

OK - so you have two pairs of resistors in parallel - the pairs are in series, and there is a lone resistor also in series.

Some tips:
You need to make the circuit so it's all one kind of combination ... you do this by combining resistors in clusters.
Here the obvious approach is to tun this into three resistors in series.
You may get circuits where it is easier to turn everything into parallel.

The parallel combination rule is the one with the fractions: $$\frac{1}{R_{tot}}=\frac{1}{R_1}+\frac{1}{R_2}$$ ... the rest is a bit of algebra.

The next thing to realize is that, in a series combination, it does not matter which order the resistors come in. The lone resistor does not have to be between the two sets of parallel resistors.

The image does not actually show any of your work ... just a table with some results on it.
What you should do is work through the example one step at a time here and we will be able to see how you are thinking - and tweak your approach to ace that exam.

3. Feb 3, 2014

### chococat

Ok, I understand.
I attempted it again this morning, and was able to find a good website with a good example to work through on my own. I think the problem that I was having was that when I went to find the current for the resistors in parallel, I was getting a higher number than my overall current.
Also, I was having issues with the voltage drop for resistors in parallel.
I think I understand what I was doing wrong- I didn't realize that voltage drop is the same in resistors in parallel, and how I can solve for it & also use it to find my current.
I solved for all the blanks in the table and showed my work too.
I think I got it all right, but if you can double check it, that would be great.
http://i.imgur.com/DCdR6ZF.jpg?1?1905

4. Feb 4, 2014

### Simon Bridge

Yeah - the "voltage" is actually the property of a point in the circuit rather than the components. If you put the negative terminal of a voltmeter on the ground or negative terminal of the battery, you can use the positive terminal to trace the way the voltage changes as the current goes around.

The voltage across a component is actually the difference between the voltages on either side of the component, which is why you'll see it called the "potential difference".

With the resistors in parallel, the points on either side of one of the resistors are the same as those for the other one ... so the voltage across them is the same.

You can also think of it as a stream flowing downhill - in which case the height is the voltage.

5. Feb 5, 2014

### lightgrav

what helps most is actually drawing each simplified circuit.
I think it is best to do them below the original, in order.
you can organize your work to do the simplification steps on the left of the diagrams, all the way to the bottom, then back-substitute from the bottom right to the top right, near the appropriate "effective resistor" .

"when you find a new current, you can calculate a new voltage".
(and a new voltage allows new current to be calculated)

6. Feb 5, 2014

### ehild

It is all right, except the power on the 8 ohm resistor.

ehild