Here's the circuit diagram for a current to voltage proofing

enter image description here

and here is the proof for getting \$\dfrac{V_o}{i_s}\$


Question is:

How did it ended up getting the \$\large-R_1\left({1+\dfrac{R_3}{R_1}+\dfrac{R_3}{R_2}}\right)\$?

It just seems mind blowing and things start to get jumbled. Getting the relationship is easier if \$R_2\$ and \$R_3\$ are not included, but the diagram includes them.

  • \$\begingroup\$ This is a very irresponsible and misleading schematic. Junction dots are important, whether you think they are necessary or not. Always put a junction dot where two schematic lines are meant to show a connection. At a T connection, it's clear, but at a cross it's otherwise not clear, especially if there are no junction dots elsewhere as in your schematic. It looks like the bottom of Is and the right side of R2 are supposed to be connected to ground, but your schematic says otherwise. Attention to detail matters. -1 for sloppiness and thinking you are above the standard conventions. \$\endgroup\$ Aug 28, 2012 at 12:10
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    \$\begingroup\$ @Olin - I guess you should tell that to the publisher; it looks like it's from a textbook. (In which case all schematics in the book will be like that :-)) \$\endgroup\$
    – stevenvh
    Aug 28, 2012 at 13:10
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    \$\begingroup\$ @stevenvh: No, it's not my issue where the OP got the schematic from. I don't care if he drew it himself on a napkin, used schematic software himself, or copied it from elsewhere. He is the one that posted it here, so he has the responsibility for it. There is lots of crap on the internet. It is in no way acceptable to post crap just because you saw it elsewhere. \$\endgroup\$ Aug 28, 2012 at 22:50
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    \$\begingroup\$ I agree with Olin Lathrop. If the cross at the bottom is no connection then the problem becomes a completely different one \$\endgroup\$ Jul 29, 2013 at 13:19

2 Answers 2


That's from a textbook, right? They make it complicated so that you have to apply the things you've learned to solve the problem. With R3 = 0 and without R2 It would be simply

\$ V_O = I_S \times R_1 \$

Now you have to apply KCL to the node where the 3 resistors meet. Let's call the voltage there \$V_1\$. Then we get

\$ \begin{cases} V_1 = R_1 I_S \\ V_1 = R_2 I_{R2} \\ V_O - V_1 = R_3 I_{R3} \\ I_S + I_{R3} = I_{R2} \end{cases} \$

The first three equations define the voltages across the resistors, the last on is KCL. The first equation gets you \$ V_1\$ directly. So we're left with a set of 3 linear equations in 3 unknowns (\$I_{R2}\$, \$I_{R3}\$ and \$ V_O\$), which is easy to solve with a couple of substitutions.


I have struggled with the same problem. While searching about it, i found this topic and then solved the problem. Here is the solution. v' is the common node of 3 resistors. enter image description here


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