# Source Conversion: Combining series current sources

I have done 1.1 and gotten 9V and 3k Ohms for the equivalent source. I am trying to do 1.2 without first converting to voltage source and I come across this situation:

How would I combine the two current sources that I have circled into one current source, without converting them first to their equivalent voltage sources.

• How did you get $3\:\text{k}\Omega$ for your equivalent source impedance in step 1.1? – jonk May 21 '19 at 17:38
• I just used Vs = Is * Rs to convert the top left current source into a voltage source. Then I combined the two series voltage sources on the left and the series resistances. I also used Vs = Is * Rs to convert the middle Current source into a voltage source. Now having two voltage sources in parallel with resistors, I found that to be equivalent to a 9V source. Then I added R4 and R11 to get the series resistance of 3k ohms. – mk3009hppw May 21 '19 at 17:46
• I get the $9\:\text{V}$ the same as you. But I don't get the same output impedance taking the dashed lines as the boundary, anyway. Suppose you short $a$ to $b$ (labeling the top point $a$ and the bottom point $b$.) What current would you see in $R_4$? – jonk May 21 '19 at 17:47
• Shouldn't it be just adding R4 and R11 to get the output impedance? – mk3009hppw May 21 '19 at 17:49
• No. You missed something important. – jonk May 21 '19 at 17:49

The transformations I see are shown below. Just follow the red arrows:

simulate this circuit – Schematic created using CircuitLab

The last one converts to:

simulate this circuit

So what is the source impedance, again?

• Oh okay. So it should be simpler to do If I convert them at the end to current sources so that I can just add the current sources in parallel. – mk3009hppw May 21 '19 at 18:15
• @mk3009hppw So long as the logic I used makes sense to you, that's all that is important. Does it make sense? (And although there are several ways to approach this, I chose a way that I hoped would be easy to follow and you'd agree with, easily.) – jonk May 21 '19 at 18:17

It would be simpler to solve this by replacing I1 and R2 with a thevenin equivalent, instead of replacing V1 and R1 with a norton equivalent.

Then you'll have to switch to norton to combine with I2 and R3, then back to thevenin to get the final answer.

If you line up the sources and resistors, you can see that the current going into the resistors is 8.3mA. Next, you can calculate each resistor current using the current divider equation, getting i_1k = (3k/(1k+3k))*(5mA + 3.3mA). Finally, you can use KCL to find the output current at the top node as i = i_1k - 5mA. My oppologies if this is wrong, but my sticky note calculation is getting 1.225mA.

simulate this circuit – Schematic created using CircuitLab

Then the equivalent resistance can be found to determine the Norton equivalent circuit by opening the sources and evaluating the net resistance (4k Ohms).