# How can I solve a circuit of resistances with both a voltage and a current-source in between them?

If I have a circuit that consists out of multiple non-parallel/serial resistances with current/voltage sources in between them like the one below

How can I solve a system like that (in this case for $I_{10V}$ and $V_0$)?. Would I have to do on a Thévenin equivalent or is there a simpler way?

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1. Treat the current source as an open circut and solve for Vo.
2. Treat the voltage source as a short circuit and solve for Vo.
3. Add the Vo from 1 and 2. This is summing the contribution from the current source and the voltage source and is the output voltage of the circuit.
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simulate this circuit – Schematic created using CircuitLab

Note that the R1 is a resistor in series with a current source, I1. A resistor in series with a current source can be removed because it does nothing: 100 mA of current will flow whether you short that resistor, or whether you replace it with a 5 $M\Omega$ resistor.

Therefore, let us make the simplification:

simulate this circuit

Now we have a resistor in parallel with a current source. That can be replaced by a voltage source in series with a resistor. Imagine I1 and R4 disconnected from the rest of the circuit. If I1 and R4 are unloaded, R4 develops 1V. So the Thévenin voltage is 1V. If I1 and R4 are short-circuited, then all of the 100 mA flows through the short load. So this means that the Thévenin voltage of 1V must be in series with a 10$\Omega$ resistor, which limits the current t

simulate this circuit

Now you have only resistances and voltages in the circuit, which makes it simpler to figure out the Thévenin voltage at Vo. The Thévenin resistance is obtained by replacing V1 and V2 with short circuits and solving the remaining resistor network.

A good way to proceed here is to realize that the order of R6 and V1 doesn't matter: two elements in series can be commuted. Then when V1 is next to V2, they can be combined into a single voltage source.

simulate this circuit

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Either your answer or mine is incorrect, because they give different answers. Either that, or I made a mistake when evaluating one of the answers... – trav1s Jun 23 '13 at 0:09
@trav1s One data point is that CircuitLab's DC solver reports the same voltage for Vo in each circuit. – Kaz Jun 23 '13 at 0:45
Ok good, I must have made a mistake trying to solve the circuits in my head. – trav1s Jun 23 '13 at 0:46