Hello, I was just hoping someone could do a sanity check on the attached question. Using Kirchoffs voltage law on the top part of the circuit, it seems there must be 27V across the 9 ohm resistor which would put -15V across the 3 ohm resistor. This means 5A for I2 is flowing into the node. When we add this to the 3A of I1 this will give us a value of 8A for I3. Does this seem to work out? many thanks Jim
Thanks, my answer just does not seem to sit right with me. As we are missing two bits of information, (R) and (E) my answer seems a bit simplistic.
I got the same, I2=5A, I3=8A. There are two reasons that we can find these solutions. 1. Parallel branches have the same voltages across them. In this case the branch that contains 12 Volt source and 9 Ohm resistor also gives us current though that specific branch. So for that branch we have ALL the information. We have voltage across it and current though it. This gives us voltage across the branch that contains 3 Ohm resistor. Why? Because it is in parallel with the other branch. So we know that there is 15 Volts across 3 Ohm resistor. Apply Ohm's Law, solve for I2. 2. Parallel branches form current divider. That is what you are seeing, 3A is entering a node, from that node there are two exits/branches, so 3A is divided between those branches. Since there is one input branch and two output branches, you have three currents. Since currents either add up or subtract, all you have to do is figure out which currents are entering and which currents are leaving the node. It is kinda interesting illustration of the properties of parallel circuits.
Thank you for taking the time to respond, I am still having a bit of trouble getting my head around it. To me it seems the 3 ohm resistor is not in parallel with the 9 ohm resistor because it has a different voltage source at one end. If it was above the 9 ohm resistor in the drawing then it would make sense and we could say it had the same voltage across it. I just built a similar circuit on the bench and could not then calculate the measured values from just what was given in the question. The second EMF seems to affect the result.
You are correct. The resistors are not in parallel. The branches are in parallel. 3 Ohm resistor forms one branch. 12 Volt source (battery) together with 9 Ohm resistor form second branch. The two branches are in parallel. So you have -12 Volt + 9 Ohm * 3 Ampere = 3 Ohm * (-I2) -12+27=3*(-I2) -I2=(-12+27)/3=15/3=5 Ampere I2=-5 Ampere the minus sign indicates that the current should not be going from right to left, it should be going from left to right. Which brings us to the I3. 3 A and 5 A enter the node, I3 leaves the node. So I3 must be the sum of 3 A and 5 A, I3=3+5=8 A.
Let's just assume R is 1Ω, if 8A is flowing through it would create a PD of 8V. So what's the voltage of E? Allen
Thanks again, I get it now, I just rebuilt the circuit and what do you know, it worked! Real world and theory agree The teacher did say not to overthink it but he said every year students get stuck on it.
Hi Allen, Well you must have guessed that was part 2 of the question, except they have inserted a 20 ohm resistor for R3. Regarding your question I would say it is: R3 8V + R2 15V = 23V for B2 Sound like I am on the right track? cheers Jim
The second voltage does affect the result. The key is to note that the original question never claimed that I1 was 3A independent of E or R, merely that with, for whatever the value of E and R happen to be, that I1 happens to be 3A under those conditions. You have enough information to determine the relationship between E and R so that given one you could find the other, but you don't have enough to actually solve for them as is.