Electronics - Parallel Circuits - Discussion
Discussion Forum : Parallel Circuits - General Questions (Q.No. 4)
4.
What is the total power loss if 2 k
and 1 k parallel-connected resistors have an IT of 3 mA?
and 1 k parallel-connected resistors have an IT of 3 mA?6 
W
W36 W
W6 mW
36 mW
Discussion:
39 comments Page 4 of 4.
Piyush suthar said:
1 decade ago
Thats simple 1/r=1/2+1/1
r=2/3
p=i^2*r
=(3ma)^2*2/3
=9ma*2/3
=6mw
r=2/3
p=i^2*r
=(3ma)^2*2/3
=9ma*2/3
=6mw
Gautam said:
1 decade ago
Here r=2000/3
& i^2=(3*10^-3)^2=9*10^-6
we know P=i^2*r=6*10^-3=6mW
& i^2=(3*10^-3)^2=9*10^-6
we know P=i^2*r=6*10^-3=6mW
Chandrashekhar Manekar said:
9 years ago
Actually power( P) = I^2(R).
By using this, we can solve the problem.
By using this, we can solve the problem.
GOPI said:
1 decade ago
P=I^2*R
R=(2/3)*10^-3
I=3*10^-3
P=(3*10^-3)(3*10^-3)(2/3)*10^-3
P=6w
R=(2/3)*10^-3
I=3*10^-3
P=(3*10^-3)(3*10^-3)(2/3)*10^-3
P=6w
Syed said:
1 decade ago
I agree with ramprabhu answer.
Opion "A" is correct.
Opion "A" is correct.
BK ROUT said:
1 decade ago
Vinays answer is right. So option C is correct.
Nandini said:
1 decade ago
Yes, I also agree vit rambabu.
Anjali.r said:
1 decade ago
P=i^2*r
P=3^2*(2/3)
P=6mw
P=3^2*(2/3)
P=6mw
Abhishek gupta said:
1 decade ago
1*2=1/2*3=2*3=6W
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