Electronics - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 12)
12.
What are the branch currents I2 and I3?
Discussion:
10 comments Page 1 of 1.
Madhu said:
1 decade ago
I2 is 9.56mA
Mike b said:
1 decade ago
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??
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:
1 decade ago
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
ie 4.029mA, tat means always I2 or I3 may be less than or equal to 4.029mA
Prof. X said:
1 decade ago
R2 & R3 = 4687.5 ohm's plus R1 = 8687.5 ohms
It = 4ma
so I2 = 2.5ma
I3 = 1.5ma
It = 4ma
so I2 = 2.5ma
I3 = 1.5ma
Carlo Gonzales said:
1 decade ago
I2 = 7/2780 A.
I3 = 21/13900 A.
Why has it been multiplied by 2?
I3 = 21/13900 A.
Why has it been multiplied by 2?
Mukesh said:
1 decade ago
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.
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:
1 decade ago
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.
Req(parallel) = 4.65.
So I2 = 4.65/7.5 X 4.02mA = 2.49mA.
I3 = Itotal-I2 = 4.02-2.5 = 1.5mA.
(2)
Blueknight802 said:
10 years ago
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.
(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.
(1)
Bhargavi said:
7 years ago
Why we are taking v2 and v3 as zeros?
Anomie said:
3 years ago
I2=V2/R2
i2= 35/7.5k
i2= 4.66 mA
i3= 35/12.5k
i3= 2.8mA
Is this correct? why can't we do it like this?
i2= 35/7.5k
i2= 4.66 mA
i3= 35/12.5k
i3= 2.8mA
Is this correct? why can't we do it like this?
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