# Finding an unknown resistance with superposition

I've been given a problem in which I need to calculate the resistance of an unknown resister $R_2$ by using the superposition method. The only other info which I have is that $R_2$ needs to be such so that $I_1=I_2$ and $I_3=20\text{ mA}$.

I hope the image is readable:

I've set up the three superpositon cases, but at the start of trying to calculate the solution I realised that I had no idea how to proceed with an unkown resistor.

A sort of idea I had was to solve for $R_2$ in the first two cases, but then I don't know what to do with the third case.

• Superposition is by far the most complicated method to solve that! Do you really have to go that way? – carloc Dec 9 '16 at 21:56
• Sadly, yes. The problem specifically asks us to use superposition. But I would also be interested in other approaches. – Andy Grey Dec 9 '16 at 21:59

With superposition just write three components of I1 (say I1.1(R2), I1.2(R2) and I1.3(R2) ) each with one single source activated.

Each expression will be function of unknown R2. Then you sum them up and finally solve this to be 10mA.

I2 will automatically be 10mA too for a simple KCL at center node.

Alternatively Ohm's law would solve this right away:

Given I1=10mA Ux at center node will be $U_\text{x}=U_1-R_1I_1=6.53\,\text{V}-458\,\Omega\times10\,\text{mA}=1.95\,\text{V}$

On the other hand $$I_2=\frac{U_2-U_\text{x}}{R_2+R_\text{i}} \quad\Rightarrow\quad R_2=\frac{U_2-U_\text{x}}{I_2}-R_\text{i}=\frac{8.5\,\text{V}-1.95\,\text{V}}{10\,\text{mA}}-0.4\,\Omega\approx 655\,\Omega$$

• Fantastic! Thanks! I'm still pretty poor when it comes to applying Ohm's law (taking an introductory course) but glad that I was at least on the right track for the solution. – Andy Grey Dec 9 '16 at 22:43
• By doing $$10mA = \frac{U_1-U_2}{R_1+R_2+R_i}+\frac{\frac{1}{R_1}+\frac{1}{R_2+R_i}}{R_1}\cdot I_3$$ and solving for $R_2$ am I on the right track then? – Andy Grey Dec 10 '16 at 12:36
• Have a look to I3 term, there's something wrong, it is dimensionally wrong, you have A/ohm^2 which is not a current. Maybe thinking of current divider helps you... – carloc Dec 10 '16 at 13:22
• Not yet it's the opposite and rather complicated way too. I just remember the current divider rule ..times the other, over the sum.., so in your case $\ldots +\frac{R_2+R_\text{i}}{R_1+R_2+R_\text{i}}I_3$ – carloc Dec 10 '16 at 13:43
• I am afraid this is getting a little too long to be hosted in"comments", however we have: $$I_1=I_3\frac{R_2+R_i}{R_1+R_2+R_i}$$ and $$I_2=I_3\frac{R_1}{R_1+R_2+R_i}$$ then might discuss of why and how on a different question. – carloc Dec 11 '16 at 8:28