Electrical Engineering - Series-Parallel Circuits - Discussion

One doubt in question they asked to find the current through 15 K resistor. As we know that current in series circuit are same. So by ohm's law we can find that I = V/R.

So I = 75/15K is 5 ma right? Answer any one?

RT = 12+15+22 = 49.
RT of parallel = (10x10) /10+10) = 5
So, 49+5 =5 4
I = V/R. = 75/54 = 1.388888889 round off so the total current is 1.4 mA.

49*75/20+75=38.68
THEN
38/38.68=1.37...SO ROUND OFF 1.4

(12+15+22)+ ((10*10)/(10+10))=54kohm
I=V/R= 75/54= 1.3888
approx 1.4

But Question is Current through the 15 k resistor is approximately?

Here all of you give the answer of full resistance of the circuit not for particular one.

Ok but they add total resistance of the the circuit. And in series circuit current same in all resistance so answer is 1.4mA.

Using current divider rule after total current flow we get the answer.

RKM , we do not have any current source so how can CDR be used ?

If the question about any 10 k ohm , then we find the total resistor then use I=V/R(total) <<<= here I is as I source.

Then back to the original drawing and use CDR:

I(throught 10 kohm) = (R(total of two 10kohm)/R10kohm+R10kohm)*Isource.


I = (5/10)*1.388m.

=0.694mA for each R10 k ohm.

By using kirshof current low ( KCL ) we prove that Isource = I1 + I2.

I = (0.694 + 0.694 ) mA.

= 1.388 mA.

@Tisha's method is basic method in electrical.

5mA is correct.

Can anyone clarify it?