# Voltage v2 seems to be opposite for equivalent circuit

I'm working on exercise 3.16 in the book Foundations of Analog and Digital Electronic Circuits.

I want to know how the upper one can be converted to the bottom one. I want to get some tips on drawing the equivalent circuit and, especially for this question, voltage source v2 seems to be opposite to me because the current from v2 will flow through the - of v1.

Is the solution wrong?

• The redraw sure looks wrong. Both (-) of the voltage sources are tied together in the redraw. Not so in the original. Commented Mar 23 at 2:16

The (highly profound) solution to make the circuits node and mesh equivalent would be to flip the polarity on V2.

Tips to draw circuits with the sources high potential (polarity) on top. Signal flow generally goes left to right. Current sources generally pump the current left to right and from bottom to top (assuming no other compromise has to be made that takes precedence). "Diagonal" features and connections (current source in bottom circuit) should be avoided until needed to do so. There are always going to be exceptions and everyone is going to have their opinions on schematic aesthetics (I know I have my preferences and biases), but one thing all engineers will agree on is that a well drawn schematic should be able to be "read" like a short story in a single pass. And if done properly, it should be able to be redrawn by the same engineer, from memory, with near perfect accuracy. This may sound absurd, but will be realized with experience after studying several thousand schematics, app notes, and white papers.

The redrawn circuit you show looks wrong to me because both (-) terminals of the two voltage sources are tied together in the redrawn schematic. However, that isn't true for the original schematic. So the redrawn one is a mistake.

I'd start here, by assigning a ground node:

The two green arrows point to nodes that, if the red node is assigned ground, will have known values of $$\v_2\$$ and $$\v_2+v_1\$$. This is a good thing as it helps to greatly simplify the schematic:

simulate this circuit – Schematic created using CircuitLab

It's easy to see that there are really two separate circuits to analyze, independently. So the left-hand circuit can be separately analyzed.

The only impact of the right-hand side's current source is to modify the currents in the voltage supplies. So once you work out the details on the left-hand side, and if you need the currents for the voltage supplies, then you can add the right-hand side's value to whatever is worked out on the left-hand side, first.

• Then can I get the voltage difference penetrating resister r2 with only the left circuit? Commented Mar 23 at 3:39
• @weycdicdib The current source has zero impact on $v_x$. It cannot affect $v_x$. So you can get the voltage difference across $R_2$ without considering the current source. Commented Mar 23 at 3:40
• Do i need to use superposition to know current source doesn't have an impact on R2? ( ie. if r1, r2 being short the current will just go around the big triangle) Commented Mar 23 at 4:00
• @weycdicdib Well, I just redrew the circuit and it should be clear from that, alone. But yes, the current source simply short-circuits through the two voltage supplies, which have zero impedance and therefore the two voltage supplies entirely bypass the resistors just like a wire would. Redrawing the circuit helps to see this as two separate circuits, though. That's the better way to see this. Trying to trace out bypassed currents in more complex circuits can be more of a pain. Keeping it simple, as in these two isolated circuits, helps improve other more complex circuits for analysis. Commented Mar 23 at 4:13

The electrical network can be manipulated as you wish, provided that the topology must not vary, as in the figure:

Here is the method that I usually use to check if two circuits match or to redraw a circuit. First, assign a name to every node in the circuit. Then, draw the same set of nodes next to the circuit and add every branch between nodes in the original circuit to the new circuit.