Possibly, but the basic method never fails.You guys are over-thinking this.
Look very carefully and eventually you will spot a glaringly simple solution.
I will not give the solution away, as I would guess that the whole purpose of the question is to see who spots the easy way.
This is not real life. The 10V supply can output current or take it in. It is a perfect 10V with 0 ohm resistance. Both supplies force their voltage on the world. Does that help?If I build this in real life, the 16V supply combats the 10V supply (and possible damages it)
I get that there are some assumptions, but we can't ignore the 16V power supply in the calculations. If I3 has 1 amp because of (10V/10ohm), then there is no return current in the 16V supply. lf we calculate them together (but separately,) then 16V/12ohm resistors in series gives 1.33Amp through the resistors and 10V gives 1 amp, but that makes 2.3amps going through I2. (2.3amps x 4ohm) means 9.2V across 4 ohm resistor and (2.3amps x 6ohm resistor) is 13.8V across the 6 ohm resistor and somehow we get 23V across the 10ohms worth of resistors. I mean, does that sound like the numbers I'm supposed to get? since neither power supply is even close to that?This is not real life. The 10V supply can output current or take it in. It is a perfect 10V with 0 ohm resistance. Both supplies force their voltage on the world. Does that help?
Think about 6 + 4 ohms is 10 ohms. Next, 10 volts is forced across the 10 ohms. (don't over think! don't ask where the current is coming from!) 10V, 10 ohms, what is the current in the resistors?
The problem with that is having a positive voltage and negative current.Well, the 10v source across r2+r3 is fixed at 10v, right? That means the voltage drop across R1 must be 6v, right?
which means the current thigh R1 (2 ohms) must be 3 amps = I1, right?
Now the fun part, if the voltage drop across R2 + R3 is 10v, and the resistance is 10 ohms, then I2 = 1 amp. That means I3 must be -2 amps.
And so, they are all whole numbers, and the signs don't matter, so long as KVL and KCL are satisfied, and they are.I should also add that for all of the items requested (V1, V2, V3, and I3) the answers were in multiple choice form and were whole numbers.
So will you please tell me what you're getting for the numbers?And so, they are all whole numbers, and the signs don't matter, so long as KVL and KCL are satisfied, and they are.
16, 10, 6 are the voltagesSo will you please tell me what you're getting for the numbers?
So which resistors have those voltages across them?16, 10, 6 are the voltages
-3, 1, 1, 2, -3 are the currents
... at certain nodes; but 6, 4, 6 are the requested voltages (as MrChips confirms).16, 10, 6 are the voltages
Then I guess all battery chargers cannot possibly work, right?The problem with that is having a positive voltage and negative current.
by Duane Benson
by Duane Benson
by Robert Keim