Engineering DC Circuit Analysis & Thevenin Problem

[elecstudent1]

Homework Statement

Need to establish a Thevenin equivalent for the circuit attached.

Homework Equations

None given. Need to use Kirchhoff.

The Attempt at a Solution

I worked out the following:
I said that R3 and R4 are in series as there is no current splitting because 0A flows into the open-circuit.

Then I said that R2 is // to (R3+R4) and modeled that as a new resistor R6 with a value of 2.25 k-ohms. I then said that R1 is in series with this R6 and got a new R7 of 3.25 k-ohms. I then worked out the current is 3.7 mA.

I then wrote that the Vth (Thevenin Voltage) is equal to the p.d. over R4 relative to ground which was 3.7 mA * 5 k-ohm which ends up around 18 V so it's clearly wrong.

Any advice please?

 

[gneill]

Homework Statement

Need to establish a Thevenin equivalent for the circuit attached.

Homework Equations

None given. Need to use Kirchhoff.

The Attempt at a Solution

I worked out the following:
I said that R3 and R4 are in series as there is no current splitting because 0A flows into the open-circuit.

Then I said that R2 is // to (R3+R4) and modeled that as a new resistor R6 with a value of 2.25 k-ohms. I then said that R1 is in series with this R6 and got a new R7 of 3.25 k-ohms. I then worked out the current is 3.7 mA.

I then wrote that the Vth (Thevenin Voltage) is equal to the p.d. over R4 relative to ground which was 3.7 mA * 5 k-ohm which ends up around 18 V so it's clearly wrong.

Any advice please?

Work from left to right (from the voltage source towards the load) rather than the other way as you've done. Since the load is connected across R4 it shouldn't be a series element of the Thevenin resistance...

One way to approach it is to work from left to right making intermediate Thevenin equivalents as you "swallow up" successive components.

 

[elecstudent1]

Ok thanks. I tried that and I got a p.d. across R2 of 9V which I used as a "new" power source for the 2nd loop.

After doing the 2nd loop, I ended up with 5V as my Thevenin voltage. Is that right?

 

[gneill]

Ok thanks. I tried that and I got a p.d. across R2 of 9V which I used as a "new" power source for the 2nd loop.

After doing the 2nd loop, I ended up with 5V as my Thevenin voltage. Is that right?

Not quite. I think you've forgotten to include the Thevenin resistance of the first Thevenin "slice" in the calculations of the next one; remember that a Thevenin model consists of both a voltage source and a series resistance...

 

[elecstudent1]

Sorry for the late reply but just letting you know I figured it out. It's actually very straightforward and I was just overcomplicating it for myself (as usual). Thanks for your help!