# need help of transformation theorems (urgent)

Discussion in 'Homework Help' started by holybrings, Apr 4, 2007.

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1. ### holybrings Thread Starter Active Member

Apr 4, 2007
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can any1 help me about transformation theorems?
how to find the i in the circuit of the doc?

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2. ### hgmjr Moderator

Jan 28, 2005
9,030
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Can you be more specific?

hgmjr

3. ### holybrings Thread Starter Active Member

Apr 4, 2007
64
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i had attached a doc which contain the question i want to ask...thx

4. ### hgmjr Moderator

Jan 28, 2005
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The current source has no label indicating a value. Is that on purpose?

hgmjr

5. ### holybrings Thread Starter Active Member

Apr 4, 2007
64
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sry that miss out the current...this question didnt label any voltage

6. ### hgmjr Moderator

Jan 28, 2005
9,030
214
This is a fairly straightforward problem involving Ohm's Law. Can you show us what you have done so far and where you are stuck?

hgmjr

7. ### holybrings Thread Starter Active Member

Apr 4, 2007
64
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i just start learning only. I tried many ways to find the i but cant. Forgot another thing, the question need to find v at the 4A VDCS.

8. ### hgmjr Moderator

Jan 28, 2005
9,030
214
OK.

Taking this a step at a time.

1. You have the infomation you need to compute the voltage across the 6 ohm resistor in terms of the current i. (Ohm's Law)

2. You should also know that the voltage across the 6 Ohm and 2 Ohm resistor is the same.

3. With the info from 1 and 2 above you should be able to write an equation that has the sum of the currents in each of the three legs of circuit and set it equal to the total current available from the current source.

If I give you any more information I will have done the work for you and that is not fair to you the student.

See what you can do with the hints provided and post your work here.

hgmjr

9. ### holybrings Thread Starter Active Member

Apr 4, 2007
64
0
thanks for the tips. I'll try to do it ^^

10. ### holybrings Thread Starter Active Member

Apr 4, 2007
64
0
voltage cross 6 ohm is equal to 6i right?

11. ### holybrings Thread Starter Active Member

Apr 4, 2007
64
0
3. With the info from 1 and 2 above you should be able to write an equation that has the sum of the currents in each of the three legs of circuit and set it equal to the total current available from the current source.
(this i dont understand. Im a slow learner)

12. ### hgmjr Moderator

Jan 28, 2005
9,030
214
6i is correct for the voltage across the 6 ohm resistor.

Another HINT: The voltage across the 2 ohm resistor is the same as the voltage across the 6 ohm resistor so you should be able to write the expression of the current flowing in the 2 ohm resistor on that basis.

hgmjr

13. ### holybrings Thread Starter Active Member

Apr 4, 2007
64
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the 2ohm is it 12i ?

14. ### hgmjr Moderator

Jan 28, 2005
9,030
214
You need to take a second look at that answer.

Remember you have the voltage of 6i and a resistance of 2 so apply Ohm's Law as you would with any known voltage and resistance to find the current.

hgmjr

15. ### holybrings Thread Starter Active Member

Apr 4, 2007
64
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V=IR, so I=6i/2 so, is 3i correct?

16. ### hgmjr Moderator

Jan 28, 2005
9,030
214
You got it!

Now you have the current in the three legs of the circuit expressed in i.

Sum them all together and set them equal to the input current of 4 amps and solve for i.

Note the sign of the dependent current source as you set up your equation.

hgmjr

Apr 4, 2007
64
0
4A=3i+2i+6i
4A=11i
i=0.364A
right?

18. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Nope. Not quite there yet.

Your getting the currents and voltages confused. That is understandable though.

The current in the 6 ohm resistor is i, right? The current in the 2 ohm resistor is 3i, right? The current in the dependent current source is -2i since it is flowing in the opposite direction of the other currents in your circuit.

With this in mind, take another shot at the calculation.

hgmjr

19. ### holybrings Thread Starter Active Member

Apr 4, 2007
64
0
oops...forgot the direction...should be i=0.57.

20. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Nope, You are so close though.

Remember that the sum of the current in each of the 3 legs must add up to the 4A that is being delivered to the circuit from the current source.

You have all three currents at hand. All you need to do is write the equation with the 3 currents (expressed in terms of i) and set that equal to the 4 amps and solve for i.

hgmjr

PS: