# Calculating V0 using Voltage and Current Division

• november1992
In summary: No, you don't need to add the voltages of the 8 ohm and 4 ohm resistors. You can use KCL and KVL to solve for the unknown currents and voltages at the node where the 4 ohm and two 2 ohm resistors meet. Then you can use the voltage divider formula to find the voltage at the output node, which is also the voltage at Vo.
november1992

## 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 = $v_{s}$

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?

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.I guess I added wrong in my previous answer.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

november1992 said:
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.

gneill said:
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?

november1992 said:
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.

november1992 said:
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? :
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?

november1992 said:
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?

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

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

#### Attachments

• Fig1.gif
4.4 KB · Views: 743
Does this mean the potential across the 2 ohm resistor is 0.6 volts?

november1992 said:
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.

Excellent. That looks fine

## 1. What is voltage division?

Voltage division is the process of dividing a total voltage into smaller voltages across different components in a circuit. This can be achieved by using resistors in series or parallel.

## 2. How is voltage division calculated?

Voltage division is calculated using Ohm's Law, which states that voltage (V) is equal to current (I) multiplied by resistance (R). In a series circuit, the voltage across each resistor is equal to the total voltage multiplied by the resistance of that particular resistor divided by the total resistance of the circuit. In a parallel circuit, the voltage across each branch is equal to the total voltage.

## 3. What is current division?

Current division is the process of dividing a total current into smaller currents across different branches in a parallel circuit. This can be achieved by using resistors in parallel.

## 4. How is current division calculated?

Current division is calculated using Ohm's Law, which states that current (I) is equal to voltage (V) divided by resistance (R). In a parallel circuit, the total current is equal to the sum of the currents flowing through each branch. The current in each branch is equal to the total current multiplied by the resistance of that branch divided by the total resistance of the circuit.

## 5. What is the relationship between voltage and current in a circuit?

The relationship between voltage and current in a circuit is described by Ohm's Law, which states that the current flowing through a circuit is directly proportional to the voltage and inversely proportional to the resistance. This means that as voltage increases, current also increases, and as resistance increases, current decreases.

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