# Voltage Division Question

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#### cwible

##### New Member
Hi guys,

I'm working on a problem here where I believe one of the steps uses voltage division to find the voltage across the 10ohm resistor. However, that's not how the solution found the voltage across the 10ohm resistor. I was trying to find the voltage across the 40ohm resistor first and then that voltage would be equal to the 10ohm resistor's voltage because they are in parallel. I had the voltage division equation as: V=240* (40/(12+4+40)).

What am I doing wrong?

Thanks for the help.

#### dknguyen

##### Well-Known Member
It's because you are ignoring the 10 ohm resistor. It's part of the divider too. The current in the 12 ohm and 4 ohm resistor splits up and some goes into the 40 ohm resistor and some goes into the 10 ohm resistor which means the presence of the 10 Ohm resistor affects the voltage drop in the upper resistors of the divider. You can't ignore it.

How do you not ignore it? You combine the 40 ohm and 10ohm resistor into a single equivalent resistor so you have a 3-resistors in series. Find the voltage across this equivalent resistor, and it will be equal to the voltage across the 40ohm resistor which will be equal to the 10 ohm resistor.

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#### Nigel Goodwin

##### Super Moderator
It's even simpler than that, it's a simple two resistor voltage divider (disguised, so as to confuse), the 12 and 4 ohm are in series (16 ohms), and the first resistor, the 40 and 10 are in parallel (8 ohms) and are the second resistor. Notice the 'convenient' values for the calculation 2/3 of the voltage across the 16 ohm (160V), and 1/3 across the 8 ohm (80V) - again, notice the nice round values.

#### Ylli

##### Active Member
Just to confuse things... You *can* look at just the voltage source, 12 ohm, 4 ohm and 40 ohm resistor, create the Thevenin equivalent, and then add the 10 ohms.

Thevenin voltage = 240 * 40/(12 + 4 + 40) = 171.429 volts
Thevenin resistance = (12 + 4)||40 = (16 * 40)/(16 + 40) = 11.429 ohms.

So now you have a thevenin equivalent network of a 171.429 volt source in series with an 11.429 ohm resistor.

Voltage across the 10 ohm = 171.429 * 10/(11.429 + 10) = 79.999 volts. 80 volts.

[and I'm sure you could do it with the Norton equivalent also, but I never was any good with that concept.]

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