# Voltage and Current in Complex Circuits

#### ileansideways

Joined Apr 1, 2007
3

This is a very complex problem that I really needed help with. I asked my classmates, and none of them seem to want to give ten minutes to explain to me how to do it.

Anyway, we were given a complex circuit and were asked to:

1) Find the equivilent resistance (I did), and it was 15 ohms

2) Find the total current ( I did, it was 1.667 amps )

3) And find the voltage and current at EACH RESISTOR

Now, this is the tricky part. I could find the current & voltage at the 10 ohm resistor and the 2.5 ohm resistor because they are both part of a simple series and I just had to plug them into ohms law, V = IR . HOWEVER, When it came to the parallel part of the circuit... I figured:

each resistor on each of the branches was 4.1667 Volts. (I guess this was wrong)

The reason I thought it was 4.1667 Volts was because I took the total voltage (25 V ), then subtracted the voltage of the two series resistors I already found, and therefore I had the voltage of the parallel portion of the circuit. Then, since by basic laws of a circuit the voltage of each resistor in a parallel circuit is the same, by default all of them should have been 4.1667. HOWEVER, I don't think that's true for when you get into the second tier parallel circuits within the parallel circuits, and thats what confuses me.

As for current, I have no clue, and my calculations are highly contingent on whether or not I know how to calculate the voltage.

I read the online textbook, and it gave the formulas for "voltage divider circuits" and "current divider circuits", would those apply?

Thank you.

Here's the problem. -->

edit : the resistor without a label is the 10 ohm resistor.

#### hgmjr

Joined Jan 28, 2005
9,027
You have arrived at the correct values for R(eq), I(total), and V(total). You have also correctly determined the voltage across the parallel sections of the circuit.

One way to proceed would be to take the voltage that you have calculated for the parallel sections and treat each of the parallel legs as an individual circuit with 4.166 volts applied. From there you could begin applying the same technique you have already used with the overall network and analyze each of these circuit legs.

hgmjr

#### ileansideways

Joined Apr 1, 2007
3
Thank you very much.

Alright, so for the first leg...

I got
V = 1.5001 and
I = .8334
for the 1.8 resistor

and for the 4 ohm resistor -->
V = 2.67
I = .6675

16 ohm -->
V = 2.67
I = .1669

and i just treat the rest of the branches the same way?

If so, then I think I've got it solved. Eureka.

#### hgmjr

Joined Jan 28, 2005
9,027
I think you have got it.

You should ace the test for sure.

hgmjr