Electrical Engineering - Series-Parallel Circuits - Discussion

Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 16)
16.
A 12 k resistor, a 15 k resistor, and a 22 k resistor are in series with two 10 k resistors that are in parallel. The source voltage is 75 V. Current through the 15 k resistor is approximately
14 mA
1.4 mA
5 mA
50 mA
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
16 comments Page 2 of 2.

Ahmed said:   1 decade ago
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.

RKM said:   1 decade ago
Using current divider rule after total current flow we get the answer.

Dhiraj kadh said:   1 decade ago
Ok but they add total resistance of the the circuit. And in series circuit current same in all resistance so answer is 1.4mA.

Haris Ismail said:   1 decade ago
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.

Tisha said:   1 decade ago
(12+15+22)+ ((10*10)/(10+10))=54kohm
I=V/R= 75/54= 1.3888
approx 1.4

ARUNsankar P said:   1 decade ago
49*75/20+75=38.68
THEN
38/38.68=1.37...SO ROUND OFF 1.4


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