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What did I do wrong?

  • Thread starter -EquinoX-
  • Start date
1. The problem statement, all variables and given/known data

http://img96.imageshack.us/img96/4527/wrongs.jpg [Broken]

2. Relevant equations



3. The attempt at a solution

so I found that:

6 = 2000i + 2000i
6 = 4000i
i = 3/2000

V0 = 3/2000 * 2000
= 3

so therefore the dependent source is 1 mA

and then combining both parallel resistors together I got: 4/3 k ohm

so V = 4000/3 ohm * 0.001 A

is this right??

the answer doesn't seem to match anything
 
Last edited by a moderator:
68
0
You are close. Treat the right loop as if the current is only going through 2kohm resistor. You do have an open loop, so no current will flow through the 4kohm resistor. What then can you say about the voltage drop across the 2kohm resistor?
 
so is it 2V?

well the answer key says it's -2V, I don't understand why
 
68
0
It would be 2V if the voltage v1 was flipped (positive at the top and negative at the bottom). You got your voltage drop across the 2kohm resistor to be 2V (positive at the top and negative at the bottom) which is right, but because the polarity is opposite of v1 we need to account for that. We can flip the polarity by negating the value of the voltage. That is why it is -2V. It would be the same if they showed V0 with an opposite polarity. You would record it as -3V if they had the polarity opposite to what they have now.
 
I thought if the current was the opposite of the voltage then it's negative, and here the current is the same direction as voltage...
 
68
0
You are right about the voltage and current directions. But here you are not finding the voltage across the 2kohm resistor, you are finding v1. Try placing v1, the polarity and all, on the 2kohm resistor (without changing the current direction) then record the voltage according to the voltage/current reference direction rule.


EDIT: Sorry if I am of not much help. Anyone please feel free to jump in.
 
Last edited:
thanks, that totally helps out!
 

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