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.
Rama said:
1 decade ago
We know that,
P=v*i
v=i*R
so P=i2*R
1/R=(1/2)+(1/1)
R=2/3kohm=(2*10^3)/3ohm
and i=3mA=3*10^-3A
p=(3*10^-3)^2*(2*?10^3/3)
P=6*10^-3W
P=6mW
P=v*i
v=i*R
so P=i2*R
1/R=(1/2)+(1/1)
R=2/3kohm=(2*10^3)/3ohm
and i=3mA=3*10^-3A
p=(3*10^-3)^2*(2*?10^3/3)
P=6*10^-3W
P=6mW
Nandini said:
1 decade ago
Yes, I also agree vit rambabu.
Sunil Kumar said:
1 decade ago
Parallel Connection
Req = R1*R2/R1+R2.
Req = 2000*1000/2000+1000
= (2000/3000) ohms i.e (2/3)K ohms.
P = I^2 * Req.
= (3*10^-3)^2 * (2/3) * 10^3
= 9 * 10^-6 * (2/3) * 10^3
= 6 m Watts
Req = R1*R2/R1+R2.
Req = 2000*1000/2000+1000
= (2000/3000) ohms i.e (2/3)K ohms.
P = I^2 * Req.
= (3*10^-3)^2 * (2/3) * 10^3
= 9 * 10^-6 * (2/3) * 10^3
= 6 m Watts
Vinay said:
1 decade ago
Current will divide as 2mA in 1K and 1mA in 2k resistors
so power is given by p=i^2*R
power at 2k=1^2*2
=2mw
|||'y at 1k=2^2*1
=4mw
Total power=2mw+4mw
=6mw
so power is given by p=i^2*R
power at 2k=1^2*2
=2mw
|||'y at 1k=2^2*1
=4mw
Total power=2mw+4mw
=6mw
BK ROUT said:
1 decade ago
Vinays answer is right. So option C is correct.
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
Linton said:
1 decade ago
@piyush suthar
How 1/2 + 1/1 will makes 2/3 ?
1/2 = (.5)+(1) = 1.5 = 3/2 not 2/3.
How 1/2 + 1/1 will makes 2/3 ?
1/2 = (.5)+(1) = 1.5 = 3/2 not 2/3.
Abhishek gupta said:
1 decade ago
1*2=1/2*3=2*3=6W
Jonathan said:
1 decade ago
Ramprabu is wrong, he should include 10^3 in total resistance, you'll get ((2/3)x10^3)(3mA)^2 = 6mW
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