# Question about voltage in the open circuit

I have this circuit here, and I'm trying to its Thevenin equivalent voltage and resistance, using superposition. When I "kill" the $V_{0}$ source, it becomes a wire and the voltage across the gap is the voltage across the network containing R2, R3 and R4.

I'm having a hard time understanding why that's the case.

• When you turn the voltage source down to zero, you are left with its internal resistance, which is zero ohms unless otherwise stated. The ideal current source would have infinite resistance when it's turned down to zero. – Chu Apr 16 '18 at 6:33

When I "kill" the V0 source, it becomes a wire and the voltage across the gap is the voltage across the network containing R2, R3 and R4. I'm having a hard time understanding why that's the case.

If you drew the circuit after "killing" the Vo voltage source, you would get this equivalent circuit:

simulate this circuit – Schematic created using CircuitLab

If then you applied the series and parallel transformation of the R2, R3 and R4 network, you'd get:

simulate this circuit

In this form, it is perhaps more obvious that the voltage across the gap (a-b) is the voltage across the resistor network, which can be solved using Ohm's Law.

Do bear in mind that the result is only the consequence of shutting down the Vo voltage source and hence is only due to the current source creating a voltage drop across resistors. You will have to account for the Vo's contribution to find out the voltage across the gap.