# help in transformation theorems 3

#### holybrings

Joined Apr 4, 2007
64
Find i and resistance seen by the independent current source if R=6 ohm / R=1 ohm.

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#### hgmjr

Joined Jan 28, 2005
9,027
This is a good one.

How far have you gotten with the problem on your own?

hgmjr

#### holybrings

Joined Apr 4, 2007
64
because of the cos, i totally dont know how to start. The 4 ohm current i find is 1.5i

#### hgmjr

Joined Jan 28, 2005
9,027
Can you show me the equation that you used to arrive at that current?

hgmjr

#### hgmjr

Joined Jan 28, 2005
9,027
Here is a hint.

Based on your earlier answer attempt, I believe you have already recognized that the voltage across the 6 ohm resistor is 6i volts.

Now notice that the dependent voltage source is also 6i volts and it has the same polarity as the voltage across the 6 ohm resistor. It should then be apparent that the voltage at both ends of the 4 ohm resistor in the 6 ohm case are identical. That means that the current flowing in the 4 ohm resistor is always 0 amps.

With this information you should be able to make a determination about the current and the resistance being seen by the independent current source at least for the 6 ohm case.

Knowledge is a lot like a nuclear reaction. It requires a critical mass to produce a self sustaining explosion. You are in the early stages of learning during which you are gaining the foundational knowledge you need to get you to critical mass. These execises may seem to be a drudge but they are a necessary building block to get you to that self-sustaining level. If you can come up with a way to derive some enjoyment out of tackling each problem and getting to the answer, you will reach that critical mass level much earlier in your career.

Good Luck,
hgmjr

#### holybrings

Joined Apr 4, 2007
64
i use I=V/R, I=6i/4=1.5i

#### hgmjr

Joined Jan 28, 2005
9,027
i use I=V/R, I=6i/4=1.5i
The problem with this equation is that it does not account for the existence of the dependent voltage source with its value of 6i volts.

You will see from my most recent reply how that 6i volt dependent voltage source effects the current in the 4 ohm resistor.

hgmjr

#### holybrings

Joined Apr 4, 2007
64
hgmjr said:
Now notice that the dependent voltage source is also 6i volts and it has the same polarity as the voltage across the 6 ohm resistor. It should then be apparent that the voltage at both ends of the 4 ohm resistor in the 6 ohm case are identical. That means that the current flowing in the 4 ohm resistor is always 0 amps.
(i dont understand this sentences)^^

#### hgmjr

Joined Jan 28, 2005
9,027
Let me see if I can state it more clearly.

hgmjr said:
notice that the dependent voltage source is also 6i volts
You can see that there is a dependent voltage source located at the far left and labeled with the value 6i volts. This is the voltage source I referred to in the statement above.

Also, you have correctly determined that in the 6 ohm case, the voltage dropped across the 6 ohm resistor can be written as 6 times i or 6i volts.

So far, so good.

That means that voltage applied to the 4 ohm resistor from the dependent power source is equal to the voltage being applied by the 6 ohm resistor with "i" current flowing through it in a direction that would make the 6 ohm resistor's top terminal positive and its bottom terminal negative. That means that the voltage applied to the 4 ohm resistor at the node where it connects to the 6 ohm resistor is at the same voltage as the node where the 4 ohm resistor connects to the dependent voltage source.

If you apply Ohm's Law to the 4 ohm resistor you will see that the result is a current of 0 amps. That means for all practical purposes the 4 ohm resistor as well as the dependent voltage source can be removed from consideration in determining the value of "i" and resistance seen by the driving current source in the case in which the resistor across the driving current source is 6 ohms.

This above explanation is only true of the 6 ohm case. The 1 ohm case is a different circuit and so the results will be entirely different.

By the way, it may be easier for you to work on this problem if you temporarily replace the value of the driving current source 2*cos(2t) with say the label "I" and then once you have performed all of your calculations you can come back later and plug in the actual value to arrive at the final answer.
Hopefully, I did a better job of explaining? If not, let me know where I have lost you and I will take another run at it.

hgmjr

#### holybrings

Joined Apr 4, 2007
64
Thanks. Understand.^^

#### hgmjr

Joined Jan 28, 2005
9,027
Are you ready to tackle the case where the resistor is 1 ohm rather than 6 ohms?

hgmjr