Why is my method of using Kirchoff Voltage Law wrong?

[zewei1988]

Homework Statement

The question asks for current I1 and I2.

Homework Equations

The Attempt at a Solution

I managed to get the answer using KCL at a few different nodes, but I am unsure why my first attempt was wrong.
For question 3a, I tried using KVL at the left most mesh.
My equation was: 3 * R1 + (-I1)* R2 = 0
and my answer for I1 using this method was 7.5A.


And why can't I use the following KCL equation to find I2:
KCL at node V1: I2 = 12/R4
which will give me I2 = 1.71A


And why do we only consider the small loop when calculating the power delivered by the 3A current source?


I'm quite weak in this subject, and my quiz is next tuesday so I would you all can help me out. Thanks

 

[Xerxes1986]

I suspect that the reason the first loop (the one including the 3A power source) gives you the wrong answer is because you didn't consider the voltage across the 3A power source. Obviously it has to supply SOME voltage, or else there wouldn't be a current. But we don't know what that voltage is, so it seems easier to use the other parts of the circuit and the junction law to solve the problem.

 

[zewei1988]

can anyone help?

 

[The Electrician]

For question 3a, I tried using KVL at the left most mesh.
My equation was: 3 * R1 + (-I1)* R2 = 0
and my answer for I1 using this method was 7.5A.

Xerxes 1986 answered this part. When you went around that loop you didn't take the voltage across the 3A source into acount; you implicitly made it zero. You can't do that.

And why can't I use the following KCL equation to find I2:
KCL at node V1: I2 = 12/R4
which will give me I2 = 1.71A

Some parts of your schematic are almost impossible to read. After magnifying and increasing the contrast, I can make out R1, R2 and R4, but the top right resistor looks like R6 and the branch current through it is apparently I2, but it isn't clear. If you have 12 volts across R4, then the current through it MUST be 12/R4 = 1.71 A. But, apparently, I2 is not the current in R4; it's the current to the left through the resistor that looks like R6, but might be R3. So, calculating the current in R4 has nothing to do with I2, which is in another resistor, not R4.

And why do we only consider the small loop when calculating the power delivered by the 3A current source?


I'm quite weak in this subject, and my quiz is next tuesday so I would you all can help me out. Thanks

Since the 3A current only flows in R1, we immediately know the voltage across it. Since we know the voltage V, then we can add the voltage across R1 (which is 3*R1) to the voltage V and get the voltage across the 3A source. It's because we know V that we only need to look at the small loop to get the power delivered by the 3A.