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 2 of 4.
Azhagusurya said:
1 decade ago
Power loss p = i^2/Rt.
Rt = (r1*r2)/(r1+r2).
Rt = (2*1)M/(2+1)k.
Rt = (2/3)k.
p = (3^2)u*(2/3)k.
p = (18/3)m= 6mW.
Rt = (r1*r2)/(r1+r2).
Rt = (2*1)M/(2+1)k.
Rt = (2/3)k.
p = (3^2)u*(2/3)k.
p = (18/3)m= 6mW.
Asha said:
1 decade ago
P = VI.....1.
V = IR......2.
Substitute above equation in 1,
we get P = I^2*R.
R = 2*1/2+1,
R = 2/3,
P = (3*10^-3)^2*(2/3)*10^3 ohm,
p = 6mW.
V = IR......2.
Substitute above equation in 1,
we get P = I^2*R.
R = 2*1/2+1,
R = 2/3,
P = (3*10^-3)^2*(2/3)*10^3 ohm,
p = 6mW.
Irshad k a said:
1 decade ago
R = 2/2+1/2.
R = 3/2.
R = 2/3.
R = 0.6668.
P = I2R.
P = 3*3*0.6668.
P = 6(approximately).
R = 3/2.
R = 2/3.
R = 0.6668.
P = I2R.
P = 3*3*0.6668.
P = 6(approximately).
Hims said:
1 decade ago
Rp = 2*1/2+1 = 2/3 kohm = (2/3) * 10^3 ohm.
P = i2*r.
i = 3mA = 3*10^-3A.
P = (3*10^-3)A*(3*10^-3)A* (2/3)* 10^3 ohm.
P = (3*3*2/3)W * (10^-3*10^-3*10^3).
P = (6)W*(10^-3) = 6*10^-3W.
P = 6mW.
P = i2*r.
i = 3mA = 3*10^-3A.
P = (3*10^-3)A*(3*10^-3)A* (2/3)* 10^3 ohm.
P = (3*3*2/3)W * (10^-3*10^-3*10^3).
P = (6)W*(10^-3) = 6*10^-3W.
P = 6mW.
Vinod said:
1 decade ago
parallel resistors:
1/r = 1/r1+1/r2.
1/r = 1/2+1/1.
1/r = 3/2.
r = 2/3.
p = v*i.
v = i*r.
p = i^2*r.
p = 3^2*2/3.
p = 9*2/3.
p = 6mA.
1/r = 1/r1+1/r2.
1/r = 1/2+1/1.
1/r = 3/2.
r = 2/3.
p = v*i.
v = i*r.
p = i^2*r.
p = 3^2*2/3.
p = 9*2/3.
p = 6mA.
Abc said:
1 decade ago
What would be the total power loss had the circuit been connected in series?
Prasanna said:
1 decade ago
Ramprabhu is wrong,
Resistance value here given is Kohms and he consider it to be only ohms
Resistance value here given is Kohms and he consider it to be only ohms
U.suruthi said:
1 decade ago
R = (2*1)/(2+1) = 2/3.
I = 3*10^-3.
P = I^2*R.
P = (3*10^-3)^2*(2/3).
P = 6*10^-3w.
I = 3*10^-3.
P = I^2*R.
P = (3*10^-3)^2*(2/3).
P = 6*10^-3w.
Sudhakar said:
1 decade ago
Here total current given is It=3mamp,
And R1=2kohm, R2=1kohm are in parallel,
So equivalent resistance of 2 parallel resistors is Req=R1*R2/ (R1+R2).
So Req=2kohm*1kohm/ (2kohm+1kohm).
= (2*1*kohm*kohm) / (3kohm).
=2kohm/3.
Now total power loss in a circuit is=It^2 (Req).
= (3mamp*3mamp) (2kohm/3) =6mw is the correct answer.
And R1=2kohm, R2=1kohm are in parallel,
So equivalent resistance of 2 parallel resistors is Req=R1*R2/ (R1+R2).
So Req=2kohm*1kohm/ (2kohm+1kohm).
= (2*1*kohm*kohm) / (3kohm).
=2kohm/3.
Now total power loss in a circuit is=It^2 (Req).
= (3mamp*3mamp) (2kohm/3) =6mw is the correct answer.
Anjali.r said:
1 decade ago
P=i^2*r
P=3^2*(2/3)
P=6mw
P=3^2*(2/3)
P=6mw
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