# Voltage and current division

## Homework Statement

http://i.imgur.com/XbMaK.png

Use voltage and current division to determine $V_{0}$ in the circuit given that $V_{out}$ = 0.2V.

## Homework Equations

$v_{i}$ = $\frac{R_{i}}{R_{eq}}$ * $v_{s}$
$i_{n}$ = $\frac{R_{eq}}{R_{n}}$ * $i_{s}$

## The Attempt at a Solution

I got $\frac{236}{23}$ for the $R_{eq}$

0.2 = $\frac{1}{\frac{236}{23}}$ * $v_{s}$

2.05 = $v_{s}$

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gneill
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The result doesn't look right. What are those formulas under Relevant Equations? What do Ri and Req represent? How did you calculate the Req that you used in your attempt?

phinds
Gold Member
2019 Award

Just think logically for a minute. What you have is a stream of voltage dividers. How could you possible have a source voltage that is comparable to the output voltage the way you do? Just at a glance you should see that it has to be at least an order magnitude higher

My book lists those formulas under voltage and current division.
Any individual resistor is Ri,Rn
Req is the equivalent resistance in the circuit

I started out by combining the 1 and 2 ohm resistors to make a 2/3 resistor. Then I combined that with the 2 ohm resistor in parallel to make a 1/2 ohm resistor. I combined that with the 4 ohm resistor in series to make a 4.5 ohm resistor and then combined that result with the 4 ohm resistor in parallel to make a 8.5 ohm resistor. Finally I combined the 8 ohm resistor in series to get 16.5 ohms.

16.5 = $R_{eq}$

0.2 = $\frac{1}{16.5}$ * $v_{s}$

3.3 = $v_{s}$

Last edited:
I just did what I read in my book. I found the equivalent resistance so I could find the proportionate fraction of the voltage for the 1 ohm resistor. So since 0.2v is given I solved for Vs

gneill
Mentor
My book lists those formulas under voltage and current division.
Any individual resistor is Ri,Rn
Req is the equivalent resistance in the circuit

I started out by combining the 1 and 2 ohm resistors to make a 2/3 resistor. Then I combined that with the 2 ohm resistor in parallel to make a 1/2 ohm resistor. I combined that with the 4 ohm resistor in series to make a 4.5 ohm resistor and then combined that result with the 4 ohm resistor in parallel to make a 8.5 ohm resistor. Finally I combined the 8 ohm resistor in series to get 16.5 ohms.
Ah. Unfortunately that technique will not work here. You might determine the net resistance in order to calculate the load on Vo (the diagram designates the source voltages as Vo), perhaps to determine the current it supplies as I = Vo/Req, but it won't allow you to find Vo given Vout.

I can see several possible approaches, two of which are as follows.

1. Reduce the network behind the 1 Ohm resistor to a single resistor and voltage supply. That is, find the Thevenin equivalent of the network up to but not including the 1 Ohm resistor, leaving the Vo as an unknown. Then apply the voltage divider formula to the resulting simplified circuit to extract a value for Vo. Reduction of the circuit can be accomplished stepwise by working from left to right, forming successive Thevenin models as you go.

2. Starting at the output, note that with Vout specified you can find the current in the 1 Ohm resistor. Where does that current come from? It must come from the 2 Ohm resistor. So what's the voltage drop across that resistor? What then is the potential at the next junction (where the 4 Ohm resistor meets the two 2 Ohm resistors)? Continue to work back in a similar manner until you reach Vo.

phinds
Gold Member
2019 Award
2. Starting at the output, note that with Vout specified you can find the current in the 1 Ohm resistor. Where does that current come from? It must come from the 2 Ohm resistor. So what's the voltage drop across that resistor? What then is the potential at the next junction (where the 4 Ohm resistor meets the two 2 Ohm resistors)? Continue to work back in a similar manner until you reach Vo.
Yeah, that's what I did. It's VERY simple and as quick as any method you're likely to find.

The 2ohm resistor has a current of 0.2A so it's Voltage is 0.4V. At the node where the 4 ohm and two 2 ohm resistors meet is what confuses me.

If I use KCL, then I1+I2 = 0.2A. So how can I solve for the two unknown currents?

gneill
Mentor
The 2ohm resistor has a current of 0.2A so it's Voltage is 0.4V. At the node where the 4 ohm and two 2 ohm resistors meet is what confuses me.

If I use KCL, then I1+I2 = 0.2A. So how can I solve for the two unknown currents?
What's the node voltage?

I don't think I can use the node voltage method. My professor says that we can't use information we haven't learned yet to solve our current homework problems.

gneill
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I don't think I can use the node voltage method. My professor says that we can't use information we haven't learned yet to solve our current homework problems.
You can't add together two potential changes? :surprised:
If so, I don't see how you can do any problems at all!

Well, there is a section in the next chapter called the Node-Voltage method so I thought you were referring to that.

Is the voltage 0.4V?

gneill
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Well, there is a section in the next chapter called the Node-Voltage method so I thought you were referring to that.
No, just the voltage (or potential) at the node.
Is the voltage 0.4V?
That's just the potential increase as you cross over the 2Ω resistor.

You want the potential with respect to the common rail that runs along the bottom (it's a convenient reference point for potentials for this circuit).

I'm still confused, am I supposed to add the voltages of the 8 ohm resistor 4 ohm resistor to find the voltage at that node?

gneill
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No, you do a KVL "walk" from the bottom of the 1Ω resistor, through the 2Ω resistor to the node. You're building on the values that you're calculating along the way...

is this correct?

-0.2V + 0.4V +2I = 0

i = -0.2

gneill
Mentor
You just want to "walk" from the reference node to the node in question, adding up the potential changes along the way:

#### Attachments

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Does this mean the potential across the 2 ohm resistor is 0.6 volts?

gneill
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Does this mean the potential across the 2 ohm resistor is 0.6 volts?
It does indeed! So now you can work out what current it must be flowing through it. And you already know what current flows out through the other 2Ω resistor... a little KCL at the node should then tell you the current through the 4Ω resistor... carry on.

gneill
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Excellent. That looks fine