- Joined Feb 9, 2018
This is the circuit that I'm working with.
I need to find the Thevenin equivalent. That is, I need to find the Thevenin resistance (RTH) looking into the circuit from the load, and the Thevenin voltage (VTH) or the voltage across the load.
So, to find the Thevenin equivalent, I need to start by removing the load. This makes the circuit into a simple parallel circuit with the following attributes:
RT = (R1+R3)//(R2+R4) = 5.76 ohms
IT = 12.5A
I1//3 = 8A
I2//4 = 4.5A
Assuming that all looks correct, I just need to figure out how to find the voltage across the load. Since (R1+R3)//(R2+R4), they will have the same voltage across them (72V) and dividing that voltage between the resistors in each parallel branch is relatively easy. What I'm unsure of is how the load will affect this circuit. For the sake of argument, we can assume the load to be a multimeter or, more accurately, what a multimeter would see when connected to terminals Vb and Va. Would the load, then, be the voltage drop between R2 and R4 (if we assume that Va is the positive terminal)?
I have a similar issue when trying to find RTH. To find RTH, I remove the voltage source and find RT looking through the circuit from the load. The load, effectively, becomes a current source with infinite resistance (an ideal multimeter) for the sake of measuring resistance. If we assume Va to be the positive terminal, the current immediately splits into R2 and R4. Looking at the flow of current, it seems like R2 then becomes in series with R1 and R4 with R3. Is this correct? And, if so, are these sets of series resistors in parallel with one another?
Thanks in advance. I'm still new to this and am trying to get my head around how everything relates.