V = IR
Rseries = R1 + R2 + ... + Rn
Rparallel = (1/R1 + 1/R2 + ... + 1/Rn)^-1
The Attempt at a Solution
I simplified the circuit, to find I = 3mA. What I did was this (in order starting from far right):
6||3 = 2kohm, which is in series with 16kOhm, so 18kOhm.
The 18kOhm is in || with 9kOhm, so 6kOhm.
6kOhm is in || with 6kOhm, so total of 12kOhm.
The 12kOhm is in || with 4kOhm, as well as the 6kOhm. So I got 2kOhm for that part.
Then, then 1kOhm and the 2kOhm are in series, so that equals 3kOhm total equivalent resistence.
Then, I found that I = 3mA.
Now, it says to find I_0. I have no clue as to how to do this. I think my simplifying is correct.
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