sorry, then i guess i am wrong. The answer is given by my teacher.ecjohnny,
the current flow in the 1k is not 2mA. The voltage drop across that 1k is not 2V. Look at your circuit ... redraw it if necessary.
How did you come up with 3k as an answer?
ecjohnny,
I'm sure those who responded to you can explain the answer to you.
Show the work you've done so far, then someone here can see where you made your mistake.
You are correct when you stated a parallel branch has the same voltage, but in this case, the 1k and the 4k are not in parallel. To solve this problem you'll be working with KVL, KCL, and Ohms Law. If need be, for your own clarity, work the circuit down to a single equvalent resistor.
You need to read the first volume of ebooks at this site located at http://www.allaboutcircuits.com/vol_1/index.html
Then answer the questions presented by papabravo and recca02.
If we told you the answer, what did you gain? After all, your teacher told you the correct answer.
ok here is another try.
{note 4k and r are in series and this series combn is in parallel with the 1k,hence voltage across 4k +r and 1k is same.}
when the 4 ma current reaches the parallel resistances of 1k and (4k +R) it splits into two currents one through 4k +r and other through 1k.
u know current in series element is same and hence as mr papabravo pointed
out current through R will be same as current through 4k which can easily be calculated using v=ir (see mr p.b's post).
once this current is know to u. apply kcl to get remaining current through the 1k (simply subtract from 4ma in this case). now use v=ir to get voltage across 1k which is same as voltage across 4k+R.
now u know voltage across 4k+R . and the current thru it (same current which flows thru r and 4k) .apply v=ir again here.u get toal resistance of this branch.now subtract 4k to get the value of r.
try solving some parallel and series combination to get a better understanding of the subject.
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