Thevenin equivalent, of circuit with one voltage and one current source

I am a Computer Engineering Student, and we have a lesson called Electric Circuits. I am trying to understand Thevenin's Theorem, but there is a circuit that confused me.

simulate this circuit – Schematic created using CircuitLab

I removed power and current source to calculate Rth and if I am not wrong it is 1Ω. I also calculated Vth, with the voltage source back on, to be 5V.

My question is what happens with the current source? When I put it back on the circuit, I shorted the voltage source, but at the same time this shorts out the current source. So, is it going to affect the rest of the circuit, or not? How would you solve this?

Probably a silly question, but I am inexperienced with this stuff. Thanks in advance.

The current source does not affect this circuit at all.

The voltage source forces a particular voltage across the nodes the two sources share. If the resistors and the load connected between a and b do not sink 1 A with 10 V across them, then the voltage source will sink the excess current.

This of course only strictly applies to the theoretical circuit using an ideal voltage source.

Many real devices that we might crudely model with a 10 V source can only source current but not sink current (Other kinds of voltage source, like shunt regulators, can only sink current).

• Of all the answers I see, this one is squarely directed at the OP's real question about the current source, as simply as possible but without being simplistic. – jonk Nov 15 '17 at 20:04

In this circuit, the total current through the resistors = 10/4 = 2.5 Amps. If you remove the current source from this circuit, the total current through the resistors will still be 2.5 Amps. But current drawn from the voltage source is different in both cases.

In the first case, the current source provides 1 Amps. Only the rest 1.5 Amps are drawn from the voltage source. At the +ve node of the voltage source, both the currents add up and flows to the resistors ( KCL)

In the second case, the voltage source has to provide the entire 2.5 Amps.

You can use superposition theorem to easily understand what is the effect of individual sources in circuits like these.

We are assuming ideal voltage and current sources here.

For a voltage source no matter what you do with the load current the voltage does not change so its Thevenin resistance is zero (or a short).

For a current source no matter what you do with the load voltage the current does not change so its Thevenin resistance is infinite (or open circuit).

Since $R_{th} = \dfrac{\text{d}V}{\text{d}V}$

So to find our circuit $R_{th}$ we remove all current sources and replace all voltage sources with a short. When we do this we see $R_1$, $R_2$ and $R_3$ are all in parallel so $R_{th} = 1 \Omega$

The Thevenin voltage $V_{th}$is the voltage at the output with no additional load. We can replace $R_2$ and $R_3$ with $R_{23} = \dfrac{R_2 \cdot R_3}{R_2 + R_3} = 2 \Omega$

$V_{th} = \dfrac{V_1 \cdot R_{23}}{R_1+R_{23}} = 5 \text{V}$

From the outputs point of view (a to b) this is equivalent to:

simulate this circuit – Schematic created using CircuitLab

What happens to the current source: It continues to provide 1A. Without any additional load on the output the current in $R_1$ is 2.5A so $I_1$ provides 1A and $V_1$ provides 1.5A. If you add an additional load across the output $I_1$ still provides 1A but $V_1$ provides more.

In thevenin : open the current source and short the voltage source. Yes, Rth IS 1ohm. For finding Vth, there is 4ohm in circuit , Vth=5volts, current will continue to flow in between current and voltage branch or in load , you can consider either case.

• Are you sure about the Thevinin voltage? – Warren Hill Nov 15 '17 at 18:49