Discussion :: Series-Parallel Circuits - General Questions (Q.No.12)
What are the branch currents I2 and I3?
Answer: Option C
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)|
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)|
(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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