Discussion in 'Homework Help' started by SUN SHINE, Jan 21, 2017.

1. ### SUN SHINE Thread Starter New Member

Jan 6, 2017
13
0
Hi

finding the current in the 3-ohm resistor by this way isn't it wrong?? because i think that the 3-ohm resistor is connected in paralel with the the 2 and 4-ohm resistors
so i think we should first sum 4+2=6 and the equivalent parallel resistor between 6-ohm and 3 -ohm

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,868
1,364
Yes , you can use the current divider rule but the current I that is flowing through 3kΩ resistor is equal to :
I = 6mA * (4kΩ+2kΩ)/(3kΩ +(4kΩ+2kΩ)) = 4mA.

Also notice that if I = 2mA then V = 2mA *3kΩ = 6V and (4kΩ+2KΩ) current is 6V/6kΩ = 1mA ----->2mA + 1mA is not equal to 6mA.
And this is why it mus be wrong.

3. ### RBR1317 Senior Member

Nov 13, 2010
463
87
You are correct in stating that the 2KΩ & 4KΩ resistors in series make an equivalent 6KΩ resistor which is in parallel with the 3KΩ resistor.

In the general form, the current division rule is stated in terms of conductance:

In=It∙(Gn/(G1+G2+...+Gk)) which in the particular case of two parallel resistors reduces to I1=It∙R2/(R1+R2)

So your solution of i=2mA is correct. To verify the solution calculate the voltage across the 3KΩ and 6KΩ resistors which must be equal.