Electrical Engineering - Series-Parallel Circuits - Discussion

Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 8)
8.
The parallel combination of a 470 resistor and a 1.5 k resistor is in series with the parallel combination of five 1 k resistors. The source voltage is 50 V. The percentage of the load current through any single 1 k resistor is
25%
20%
100%
50%
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
22 comments Page 1 of 3.

Himanshu Yadav said:   4 years ago
Five 1k ohm parrallel R=200..
i =V/R =50/200=0.25.
i for 1k, i = 50/1000 = 1/20 = 0.05.
i %=0.05 * 100/0.25 = 20%.
(4)

Samba said:   6 years ago
Thanks @Rohit.

Richa said:   8 years ago
Please somebody give me current divider formula with explanation.

Vandana said:   8 years ago
Thank you @Meenakshi.

Don said:   8 years ago
Thanks for the explanation of the answer.

Megha said:   8 years ago
Thank you for the explanation.

Puja said:   8 years ago
You are right @Shashank.

But can you tell me what is the formula if there is not same resistance in parallel?

Satish Reddy N said:   9 years ago
@Priya,

Where does that 1/4 comes from I am not getting that?

Shashank said:   9 years ago
Actually, they ask for a percentage. So, for 51K ohms resistor 50V/200ohm = 0.25A.
Again 50/1 Kohm = 0.05A, therefore 0.05/0.25*100 = 20%.
(1)

Parneet kaur said:   10 years ago
Of course @Meena you are right.

But this simple and quick method is only applied when all the resistors are of same value. Otherwise you can't applied when resistors are of different values are given.

What is the simple method when resistors are of different values are given?


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