# How to find voltage across resistors with only current sources? Hello,

I'm trying to find the voltages across each resistor showed in the schematic, using KCL and KVL. I'm not too sure how to start the problem since there are no voltage sources and I don't know how to find the current from a current source. I was thinking of solving this through node analysis but was a bit unclear of how to do it with current sources.

For V2 (from R2) I tried (2E-3)(2) to get the voltage but the correct answer should be 2.571V.

Thank you for any help!

• Thevenin theorem would be better to start with. – Eugene Sh. May 24 '17 at 20:57
• I have not learned that yet at this point in my Circuits class. Is there no easy way to do it with kvl/kcl? – Mariankka May 24 '17 at 21:01
• What have you tried so far? Start with KCL -- you have three loops and three loop currents. If you solve for those loop currents, each one will give you the voltage across the corresponding resistor. – Dave Tweed May 24 '17 at 21:20
• "I don't know how to find the current from a current source" - it's given, that's what makes it a current source. – Alfred Centauri May 24 '17 at 21:43
• Using just superposition and Ohm's law, you can write the answers by inspection; remove I2 (replace with open circuit) and easily solve for the voltages then remove I1 and solve again. Now, just add the two solutions together and you're done. – Alfred Centauri May 24 '17 at 21:48

Here, let me set up the equations for you: simulate this circuit – Schematic created using CircuitLab

You have five unknowns — three currents and two voltages. We'll need five equations.

KCL tells us that the sum of currents into any node must be zero:

$$I_{L1} + I1 - I_{L2} = 0$$

$$I_{L2} - I2 - I_{L3} = 0$$

$$-I_{L1} - I1 + I2 + I_{L3} = 0$$

Note that the third equation tells us nothing new; it is simply a combination of the first two.

KVL tells us that the sum of voltages around a loop must be zero, which is really just another way of saying that the voltage across I1 must equal the voltage across R1, etc. We simply apply Ohm's Law to the resistors:

$$V1 = -I_{L1} \cdot R1$$

$$V1 - V2 = I_{L2} \cdot R2$$

$$V2 = I_{L3} \cdot R3$$

Now we have five independent equations that we can solve for our five unknowns.

As Alfred Centauri says, the principle of superposition makes this problem almost trivial to solve, but you may not have been exposed to that yet.

Just look at the circuit and use common sense.
$$\I_1\$$ and $$\R_1\$$ are a current source with a parallel impedance, and $$\I_2\$$ and $$\R_3\$$ idem. Both current sources with parallel impedance can be transformed into voltage sources $$\R_1I_1\$$ and $$\R_3I_2\$$ with internal resistances $$\R_1\$$ and $$\ R_3\$$ respectively.
Now the current through $$\R_2\$$, from left to right, will be the sum of the voltages divided by the sum of the resistances: $$\I_2=(R_1I_1+R_3I_2)/(R_1+R_2+R_3)\ = (8+1)/7E3 = 9/7 \, mA \$$
Then the current through $$\R_1\$$ will be $$\I_1-9/7 \, mA\ = 5/7 \, mA \$$ so that its voltage will be $$\20/7 \, V\$$ and the current through $$\R_3\$$ is $$\1-9/7 = -2/7 \, mA\$$ and the voltage across $$\R_3\$$ is $$\-2/7 \, V\$$.