Electronics - Series-Parallel Circuits - Discussion

12. 

What are the branch currents I2 and I3?

[A]. I2 = 4 mA, I3 = 2 mA
[B]. I2 = 4.5 mA, I3 = 2.5 mA
[C]. I2 = 2.5 mA, I3 = 1.5 mA
[D]. I2 = 5.5 mA, I3 = 3.5 mA

Answer: Option C

Explanation:

No answer description available for this question.

Madhu said: (Jul 23, 2011)  
I2 is 9.56mA

Mike B said: (Sep 18, 2011)  
Total current I(1):
R(2) and R(3) are together 4k7
This in series with R(1) is 8k7
I(1) = 35/8k7 = 4mA

I(2) = 4*12.5/(12.5+7.5) = 2.5mA
I(3) = 4mA - 2.5mA = 1.5mA

Or am I doing something wrong??

Visakh said: (Oct 4, 2011)  
Anyway the total current drawn from the source is 35/(4+4.69)
ie 4.029mA, tat means always I2 or I3 may be less than or equal to 4.029mA

Prof. X said: (Apr 17, 2012)  
R2 & R3 = 4687.5 ohm's plus R1 = 8687.5 ohms
It = 4ma
so I2 = 2.5ma
I3 = 1.5ma

Carlo Gonzales said: (Jul 11, 2013)  
I2 = 7/2780 A.

I3 = 21/13900 A.

Why has it been multiplied by 2?

Mukesh said: (Jul 20, 2013)  
Total resistance of the circuit is:

4K+(12.5K||7.5K) = 4K+4.6875k) = 8.6875K.

Total current = 35/8.6875K = 4.028776 mA.

I2 = 12.5K*4.028776 mA/(20K) = 2.517985 mA.

I3 = 7.5K*4.028776 mA/(20K) = 1.510791 mA.

Hence none of the options above are correct.

Junayd said: (Mar 31, 2014)  
Using CDR.
Req(parallel) = 4.65.

So I2 = 4.65/7.5 X 4.02mA = 2.49mA.

I3 = Itotal-I2 = 4.02-2.5 = 1.5mA.

Blueknight802 said: (Jan 30, 2016)  
Nodal Analysis:

(V-35)4 + (V-0)7.5 + (V-0)12.5 = 0.

V = 18.84 V.

I2 = V/7.5 = 2.518 mA.

I2 = V/12.5 = 1.511 mA.

Bhargavi said: (Jan 12, 2019)  
Why we are taking v2 and v3 as zeros?

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