Electronics - Parallel Circuits - Discussion
Discussion Forum : Parallel Circuits - General Questions (Q.No. 15)
15.
The current through R1 will be:
Discussion:
17 comments Page 2 of 2.
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...
& then after ohms law apply...
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
So the current across r1 is the difference between the 2 currents,that is 88m-40m=48mA
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
So the current in the required branch is 0.088-0.040=48mA
Chaitanya said:
1 decade ago
Vasu is correct
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.
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.
ROOPA.R said:
1 decade ago
0.088-0.040=0.048or 48mA
(I1=I2+I3)
SO I3=I1-I2
(I1=I2+I3)
SO I3=I1-I2
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
I=V/Rt
I=48/1*10^3=48mA
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