How do I make R2 the subject of this formula?

• March 22nd 2011, 11:19 AM
becca
How do I make R2 the subject of this formula?
Hi,

How would I go about making R2 the subject of the formula on the following Wiki page, it's the one under Applications - Voltage Source (Vout = ...)? I don't know how to include the formula on here. Sorry (Worried)!

Link to page - Voltage divider - Wikipedia, the free encyclopedia

Thanks,

Becca.
• March 22nd 2011, 11:37 AM
HallsofIvy
Are you referring to
$V_{\text{out}}= \frac{R_2}{R_1+ R_2}V_{\text{in}}$?

Multiply both sides by $R_1+ R_2$ to get
$V_{\text{out}}(R_1+ R_1)= V_{\text{out}}R_1+ V_{\text{out}}R_2= R_2V_{\text{in}}$

Now, get those 2 terms including [tex]R_2[tex] alone on one side by subtracting $R_2V_{\text{in}}$ and $V_{\text{out}}R_1$ from both sides:
$V_{\text{out}}R_2- R_2\V_{\text{in}}= R_2(V_{\text{out}}- V_{\text{in}})= -V_{\text{out}}R_1$
Finally divide both sides by $V_{\text{out}}- V_{\text{in}}$

$R_2= \frac{-V_{\text{out}}}{V_{\text{out}}- V_{\text{in}}}R_1= \frac{V_{\text{out}}}{V_{\text{in}}- V_{\text{out}}}R_1$.
• March 22nd 2011, 01:49 PM
becca
Quote:

Originally Posted by HallsofIvy
Are you referring to
$V_{\text{out}}= \frac{R_2}{R_1+ R_2}V_{\text{in}}$?

Multiply both sides by $R_1+ R_2$ to get
$V_{\text{out}}(R_1+ R_1)= V_{\text{out}}R_1+ V_{\text{out}}R_2= R_2V_{\text{in}}$

Now, get those 2 terms including [tex]R_2[tex] alone on one side by subtracting $R_2V_{\text{in}}$ and $V_{\text{out}}R_1$ from both sides:
$V_{\text{out}}R_2- R_2\V_{\text{in}}= R_2(V_{\text{out}}- V_{\text{in}})= -V_{\text{out}}R_1$
Finally divide both sides by $V_{\text{out}}- V_{\text{in}}$

$R_2= \frac{-V_{\text{out}}}{V_{\text{out}}- V_{\text{in}}}R_1= \frac{V_{\text{out}}}{V_{\text{in}}- V_{\text{out}}}R_1$.

Hi HallsofIvy,

Thank you for your reply, however I was referring to the very last equation on the page. The one with "R2||RL" in it.

I'm sorry I can't write it up like you do. I'm a noob at this :)

Thanks,

Becca.
• March 22nd 2011, 03:20 PM
HallsofIvy
Then I'm not sure what you are asking. That particular equation is defining " $R_2||R_L$" as meaning $\left(\frac{1}{R_2}+ \frac{1}{R_L}\right)^{-1}$ and then noting that it happens to be equal to $\frac{R_2R_L}{R_1+ R_L}$. If you tried to solve $\left(\frac{1}{R_2}+ \frac{1}{R_L}\right)^{-1}= \frac{R_2R_L}{R_1+ R_L}$ for $R_2$ everything would cancel out- it is an identity that is true for all $R_2$ and $R_L$.