Basic ciruits

its (a)

First I tried to transform the voltage to a current souce and adding the currents but that didnt work, then i tried Nodal Analysis using the two nodes. In first one I got Equation : (V1-12)/6 + V1/6 + V1-V2/6 = 0 and in the second node I got (V2-V1)/6 -6 + V2/6 =0. When I simplified the first equation I got V1= (12 + V2)/3 . Then I simplified the second equation a little and got V2-V1 = 36 and then just switched the v1 with the first equation. and get v2=24 v. and Io - V2/6 so 24/6k = 4mA but thats not one of the answers :/

>>112069
would you mind explaining how you got that please? Id really appreciate it

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56KB, 539x375px

>>112072

>>112072
>>112071
I'm going to guess (a), 3.6ma, because:

- ideal voltage sources have zero resistance
- it's not asking you to calculate the actual current, just the contribution from the current source.

This means you can completely ignore the voltage source, and treat it as a zero-ohm resistor.

This makes the LHS a resistor network with a combined resistance of (1/(1/6+1/6))+6=9 milliohms, and the RHS a resistor of 6 milliohms, and your current-divider equation, therefore, says 2.4 ma on the left, and 3.6 ma on the right.

>>112067
Open circuit the current source and calculate Io'
Then calculate Io with the current source
Io-Io' is the contribution

Io'=0.4mA

Io=-3.2mA

contribution = -3.6mA

>>112105
Finding the contribution means in Io means we can ignore the voltage source?

>>112307
>>112219 and >>112105 give the same answer, so yes.