Give \(I_1 = 1A\),as follows
Ask for \(I_s\) = ?

 

Read the first (sticky) post in this forum. You're supposed to show your attempt at solving the problem when you ask for help.

Otherwise, you're just asking someone else to do your work for you, and that's not going to help you in the long run.

 

Sorry,but I does not mean to ask somebody to do my homework.Actually,it is the course last semester.
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Yeah,I have did something to solve this problem.But the solution I got seems a little complex.I am here for a better solution.Here is my work:

0.First of all,assign the circuit reference direction to each loopback circuit.
here,we only need to assign for the loopback circuit above.We assign clockwise.

1.Then consider the loopback circuit with voltage source 3V,apply KVL to this loopback circuit,\(U_R = U = 3V, I_2 = U/R = 3V / 1\Omega = 3A\).
The direction is the same of reference direction.So I = +3A.

Actually,it hard for me to continue to descirbe it clearly without graph.-_-(later on,I will post the remain steps with graphs...)

OK.Finally we could get(\(I_s\) is the current of the remain resistance),
\(I_x + I_s = -1 \)

\(-I_x -1 + 3 = 0\)

so,\( I_s = -2 \).

 

Did you mean to use \(\pi\) in your expression?

hgmjr

 

No,its a hand-writing mistake.corrected.

 

This problem reduces to a simple two loop problem since the 1 ohm resistor that is directly across the 3V power source does not contribute anything to the value of I nor does it contribute to the value of Is.

It would make things much clearer if you would label (e.g. R1, R2, R3) the resistors in the circuit.

hgmjr

 

Greetings MichealY,

Here is the Millman's Theorem analysis of your circuit.

\({Equation\, 1:}\, \, \, \,Vx\,=\, \Large \frac{\frac{-3}{R1}-I_S}{{\frac{\,1\,}{R1}}\,+\,{\frac{\,1\,}{R2}}} \,=\, \Large \frac{\frac{-3}{1}\, -\, I_S}{{\frac{\,1\,}{1}}\,+\,{\frac{\,1\,}{1}}} \,=\, \Large \frac{-3\, -\, I_S}{{1}\,+\,{1}} \,=\, \Large \frac{-(3\, +\, I_S)}{2}\)

\({Equation\, 2:}\, \, \, \,Vx\, =\, 1\, (R1)\, -\, 3\, =\, 1\, (1)\, -\, 3\, =\, -2 \)


Plug the value for Vx from Equation 2 into Equation 1, then solve for Is.


\(-2\, =\, \Large \frac{-(3\, +\, I_S)}{2}\)

\(-4\, =\, \Large {-(3\, +\, I_S)}\)

\(4\, =\, \Large {(3\, +\, I_S)}\)

\(I_S\, =\, \Large 1\)

The value for Is is a positive 1A. That means that the current is in the direction that was assumed by the arrow.

hgmjr

 

@hgmjr
Thanks for your reply.It seems I have made a mistake.In my solution,I moved the Is path to the left of R2.Then I apply KCL and KVL to this circuit.But It seems that this modification is wrong.Another question^_^,in which condition could we move circuit path?

However,I have not learnt Milliam'sTheorm,but I think I have figured out an easiest solution.We only need to translate the current source to power source(in which condition?),then the translated power source should be \(U_s = I_s \times R_2 = I_s\) and the R2 is connected to the translated power source,so we could get,
\(3 - I_s = (R_1 + R_2) \times I_1\)

then,\(I_s = 1A.\)

michealy.