Hello all. I'm having issues with series-parallel circuits again... In the circuit attached, R1+R2 are in parallel with R3+R4 which are in parallel with R5, correct? I can do the calculations, I am just having issues with the visualizing...
Hmm, R2 is parallel with R3+R4 + BAT 3V and R1+ BAT 2V. R5 is parallel to R4 + BAT 3V "+" series connection.
Sorry, we are doing superposition theorem and I am trying to figure out what the current through R5 would be? In the schematic I drew on my notebook I took out BAT 3V in order to calculate the current through R5 with that battery out. With that battery out, is the way I said it correct? I keep getting numbers way out of whack with the books answer. Any help is appreciated.
To do superposition, you need to put a short across all voltage sources, except the one in question. Then you do it for every other source voltage. with 3V. shorted, you still have R1 in series with the rest of the network. R5 in parrallel with R4, then R3 in series with that combo, then R2 in parrallel with that combo, and R1 in series with all that.
Yes, but I am simplifying it. R1 is in series with R2, therefore, the combination of R12 is in parallel with the combination of R34 which is in parallel with R5 when I remove the 3V source with a short. Or am I wrong?
I am getting 789.238 ohms. But I think I am analyzing this the wrong way. I should always start at the branch at the end of the circuit, right?
Now do the analysis from the right side. Now start at the very right side, R5 is in parrallel to R4. Then R3 is in series with that combo. Then R2 is in parrallel to that. Then R1 is in series to all that netwoork.
Let's do a current path, current coming from the 2V. battery, flows to R5, at the same time it flows through R4, but before it reaches these two, it flows through R3, but at the same time it flows through R3 it branches through R2, and before it flows through R2 it all starts out flowing through R1. Does that help?
After that you then remove the short from the 3V. supply and put a short across the 2V. supply and then analyse the network again. Then you algebraicly add the currents through the resistor in question.
Sorry, had to finish up my homework for another class that was due now... So, I can now see how I can get the 1954 ohms. Now, I'm having a problem getting the current through R5... I keep getting pretty much 0 current through R5, which I immediately know is wrong. I have a feeling I am doing all of my math wrong.
If you've redrawn your circuit, show your work. Someone will look it over to see where you went astray.
After a lot of trial and error, I was able to get I1 = .181mA with the 3V source shorted. I can't seem to find a way to redraw the circuit with 2V source shorted in a way that will be easy to intepret. The location of the source just throws me off....
It takes practice to be able to see the network analysis. Let's start with the current path from the 3V. battery, with a short across the 2V. bat. Lets go to the extreme left side first, current flows through both R1 and R2 at the same time, but before that, it has to flow through R3 to get there, however as it is approaching R3, at the same time part of it is branching off and flowing through R5, But before any of this can happen it has to flow through R4 to begin the whole process.