Electronics - Parallel Circuits - Discussion

Discussion Forum : Parallel Circuits - General Questions (Q.No. 15)
15.
The current through R1 will be:

mcq4_1006_1.gif
128 mA
88 mA
48 mA
2.2 mA
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
17 comments Page 1 of 2.

Seshendra said:   1 decade ago
Voltage across parallel resistors are same.

So voltage across R1 is 48v,

We Know i=v/r.

=48mA.

Niraj kumar said:   1 decade ago
Total resistance Rt=(1*2*3)/(1+2+3)=1
I=V/Rt
I=48/1*10^3=48mA

ROOPA.R said:   1 decade ago
0.088-0.040=0.048or 48mA
(I1=I2+I3)
SO I3=I1-I2

Vasu said:   1 decade ago
In the answer given by Nirak kumar , Req= 1
but, its wrong. The correct formula in case of 3 r's in || is Req=R1R2R3/(R1R2+R2R3+R3R1).

So here Req will become 6/11.

Chaitanya said:   1 decade ago
Vasu is correct

Ragi G R said:   1 decade ago
By simple logic we can find it. The current in the previous branch is 0.088. This current get split into two branches. One of them is 0.040.

So the current in the required branch is 0.088-0.040=48mA

Yamini said:   1 decade ago
The current in the first loop is 88mA,in the second loop is 40mA,

So the current across r1 is the difference between the 2 currents,that is 88m-40m=48mA

Dipika said:   1 decade ago
@Niraj is correct answer becoz this three resistsnce are parellel so formula is r1*r2*r3/r1+r2+r3... so total resi r..
& then after ohms law apply...

Harika said:   1 decade ago
@according to vasu as req=6/11
I=V/Req =48*(11/6)=88mA

Shiva said:   1 decade ago
I = 0.088.
I1 = ?
I2 = 0.040.

Since I = I1+I2 (Kirchhoff current law).

I1 = I-I2 = 0.088 - 0.040 = 0.048 = 48 x 10^-3 = 48mA.


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