Calculating Vout/Vin (voltage division)

Discussion in 'Homework Help' started by TsAmE, Jun 16, 2010.

Apr 19, 2010
72
0
I am not sure how to attempt this question as Vin isnt given. If it was I would:

*Work out the first Vout after the current has gone through 1kΩ
*I would use the first Vout to work out the final Vout after the current has gone through the second 1kΩ resistor
*For these calculations I would use the voltage division formula:
Vout = Vin x R2 / R1 + R2.

How would I work out Vin, in order to do the calculations above?

File size:
1.9 KB
Views:
196
2. Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
Well assume that Vin=1V

Or we can write nodal equation an solve unknowns.

$\frac{(Vin-Va)}{R2}=\frac{Va}{R4}+\frac{(Va-Vout)}{R2}$

$\frac{(Va-Vout)}{R2}=\frac{Vout}{R1}$

But Va can be found by inspection (because we have here two voltage divider)
First
Va = ( R4||(R2+R1)/ ( R2 + R4||(R2+R1) ) * Vin

$Va=Vin *\frac{\frac{R4*(R2+R1)}{R1+R2+R4}}{R2+\frac{R4*(R2+R1)}{R1+R2+R4}}$

second

Vout = Va * R1/(R2+R1)

• 1213.PNG
File size:
4 KB
Views:
1,960
Last edited: Jun 16, 2010

Apr 19, 2010
72
0
Oh ok I see. Something else I am curious about is how the various currents flow in this circuit. Would it be like this:

*Current starts flowing at Vin (I1)
*It then goes through R2, splits at Va (the one goes through R4 to earth (I2), and the other goes through R2 (I3) therefore I1 = I2 + I3 )
*At the next common point, spilts again (one goes through R1 to earth (I4), and the other one goes through Vout (I5) (I3 = I4 + I5).

4. Markd77 Senior Member

Sep 7, 2009
2,803
594
You have to assume that no current flows out of Vout. Imagine it as a test probe point. If current flowed out and we had no way of knowing what it was, the system would be unsolvable.

TsAmE likes this.
5. hgmjr Moderator

Jan 28, 2005
9,030
214
Here is the Millman's Theorem analysis technique applied to the circuit.

I have co-opted jony130's diagram to permit comparison between the two approaches.

Begin by forming (by inspection) the Millman's Theorem equation for Vout in terms of Va:

$V_{out} \,=\, \Large \frac{\frac{V_{a}}{R_2}}{\frac{1}{R_2}+\frac{1}{R_1}}\normalsize \, \, (1)$

$V_{out} \,=\, \frac{R_1V_{a}}{R_1+R_2}\normalsize \, \, \, \, (2)$

Rearrange equation 2 to isolate Va on the left side of the equal sign:

$V_{a} \,=\, \frac{(R_1+R_2)V_{out \.}}{R_1}\normalsize \, \, (3)$

Next, form the Millman's Theorem equation (by inspection) for Va in terms of Vin:

$V_{a} \,=\, \Large \frac{\frac{V_{in}}{R_2}}{\frac{1}{R_2}+\frac{1}{R_4}+\frac{1}{(R_2+R_1)}}\normalsize \, \, \, \, (4)$

$V_{a} \,=\, \large \frac{R_4(R_2+R_1)V_{in}}{R_4(R_2+R_1)+R_2(R_2+R_1)+R_2R_4}\normalsize \, \, (5)$

Set equation (3) equal to equation (5) and solve for Vout/Vin:

$\frac{(R_1+R_2)V_{out}}{R_1} \,=\, \large \frac{R_4(R_2+R_1)V_{in}}{R_4(R_2+R_1)+R_2(R_2+R_1)+R_2R_4}\normalsize \, \, (6)$

$\frac{V_{out}}{R_1} \,=\, \large \frac{R_4V_{in}}{R_4(R_2+R_1)+R_2(R_2+R_1)+R_2R_4}\normalsize \, \, (7)$

$\frac{V_{out}}{V_{in}} \,=\, \large \frac{R_1R_4}{R_4(R_2+R_1)+R_2(R_2+R_1)+R_2R_4} \normalsize \, \, (8)$

Done.

hgmjr

TsAmE likes this.