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I'm just stuck on a question that I've been given as a quick test, it's really stumped me. I was wondering if anyone could assist me.
Here's the question:
Thank you.
Here's the question:
Thank you.
I've just started the electronic course so I'm quite new to the method that you're using. So I don't quite understand it.You can solve this via mesh analysis: write the currents I1,I2,I3 as loops passing around each of the three independent loops, the write the three mesh equations, taking proper care of the directions which you assigned (I used clockwise for all three) when summing the currents from shared resistors.
6V = 2.2k(I1-I2)
0 = 2.2k(I2-I1) + 10k(I2) + 1k(I2-I3)
0 = 1k(I3-I2) + 3.3k(I3)
Then solve the three simultaneous linear equations via your favorite method (substitution, elimination, Cramer's rule), and I3 is your answer.
I also considered a Delta-Y transform but it didn't seem to make the solution go any faster.
What do you mean by draw circuit paths?I suggested that you should draw some current paths, to help with the analysis, for a good reason.
Keep it as 3 resistors, 2.2, 10 and 0.8. Draw some current paths, these being shown as closed loops.
Oh I understand what you mean now, so once I've found the total current what formula do I use to find the current going through the resistor?The 10k and 0.8k resistors are in series; the 2.2k is NOT in parallel with the 0.8k resistor because there is something on the upper bus between them: the 10k resistor.
So combine the 10k with the 0.8k in series: 10.8k, and then you are left with two resistors in parallel, the 2.2k and the 10.8k. Combine them and then solve for the total current.
Not exactly.So once I get the current for the 10.8 resistor do I divide that current by 1and 3.3
How many volts are lost across the 10k resistor? Use Ohms Law.Im quite confused, what calculations would I need to perform?