# Find resistor.. theory help.

#### ecjohnny

Joined Jul 16, 2005
142
Guys a little help needed. Can anyone explain to me how to get the answer?
Thanks.

The answer is : 3KΩ #### Papabravo

Joined Feb 24, 2006
19,022
I'll give you a hint:
Rich (BB code):
2V/4K = 0.5 mA
So if 0.5 mA is going through the 4K, how much of the original 4 mA is going through the 1K?

#### recca02

Joined Apr 2, 2007
1,212
there are a lot of ways to solve the above problem.
u can alsoapply kvl,

one way to solve it wud be,
use current division rule according to which the current in a path is the total current into the resistance of other path divided by sum of resistance in the individual paths.
hence current thru r and 4k resistance is: (applying at the node)
I' = I*(1k)/{(r+4k)+(1K)}
hence voltage across 4k resistance will be this current multiplied by resistance.
equate to get the answer.

#### ecjohnny

Joined Jul 16, 2005
142
sorry i still dont get it. The best i know is :

1) the current flow through the 1k is 2mA. 2v/1k? parallel resistor have same voltage?

2) therefore current through R is 2mA??

3) how do i get voltage?

i might be wrong.. anymore further explanation/hints?

#### JoeJester

Joined Apr 26, 2005
4,390
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

Joined Jul 16, 2005
142
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?
sorry, then i guess i am wrong. The answer is given by my teacher.

Anyone can explain the answer to me?

#### JoeJester

Joined Apr 26, 2005
4,390
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.

#### ecjohnny

Joined Jul 16, 2005
142
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.

Actually, i knew that KVL,KCL laws and stuff. but i still dun understand.

So...Are you guys gonna help me?

#### recca02

Joined Apr 2, 2007
1,212
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.

#### ecjohnny

Joined Jul 16, 2005
142
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.
subtra

thanks.. i got it. anyway, the last part, i uses the 3.5v-2v to get voltage of R. This is KCL law am i right?

#### recca02

Joined Apr 2, 2007
1,212
kcl is kirchoff current law
summation of current entering the node(point whre many wires meet)
= summation of current leaving the node.
you subtract 0.5ma which is leaving node from 4ma entering node to get the current thru 1k which is again leaving the node.
in case of ac the summation is vector summation.

#### recca02

Joined Apr 2, 2007
1,212
that is kvl.
-voltage law
kcl -current law
states that summation of current entering a node is equal to sum of current
leaving the node.
node is the point where two or more wires meet. so that current may enter from some and leave from rest.
subtracting 0.5 ma (exiting the node) from 4ma (entering the node) will give current through the 1k resistance (leaving the node).
kcl is applicable since there is no accumulation of charges at the node number of charges entering is equal to no of charges leaving.
in case of ac summation is vector sum.

sorry i thought my earlier post didnt get posted.(pls delete the earlier one if possible)

#### JoeJester

Joined Apr 26, 2005
4,390
Yes you are correct on the voltage across R.

Follow recca's advice and practice with more circuits. It's the application of KVL and KCL where you need work.

Here's what I did last night ... before I responded to you.

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