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.
Shah qaisar said:
7 years ago
P = i^2R.
I = 3mA= 0.003A,
R = 2000 * 1000/2000+1000= 666.6 ohm,
P = (0.003 * 0.003)* 666.6= 0.00599~6mW.
P =(.003 * .003).
Option C is correct.
I = 3mA= 0.003A,
R = 2000 * 1000/2000+1000= 666.6 ohm,
P = (0.003 * 0.003)* 666.6= 0.00599~6mW.
P =(.003 * .003).
Option C is correct.
(1)
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.
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
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.
Devaki said:
1 decade ago
P=I^2(R)
R= 2K PARELLEL TO 1K OHM
= 2/3 K OHM
P= 3mA * 3mA * 2/3 * 10^3
= 6mW
SO I CONCLUDE THAT OPTION 'C' IS CORRECT
R= 2K PARELLEL TO 1K OHM
= 2/3 K OHM
P= 3mA * 3mA * 2/3 * 10^3
= 6mW
SO I CONCLUDE THAT OPTION 'C' IS CORRECT
RAMAKANT said:
1 decade ago
Rt=2/3k ohms
where we know
P=I2*R
then
P=(3*10^-3)(3*10^-3)(2/3*10^3)
P=6*10^-3w
P=6mw
THEREFORE 'C' IS THE CORRECT ANSWER
where we know
P=I2*R
then
P=(3*10^-3)(3*10^-3)(2/3*10^3)
P=6*10^-3w
P=6mw
THEREFORE 'C' IS THE CORRECT ANSWER
Krish said:
9 years ago
Two parallel resistor so equivalent resistor is 2/3 and also current 3 mA. So power loss is (p = Isquare R) 9 * 2/3 = 6 mA.
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.
Ana said:
10 years ago
1/R = 1/2+1/1.
1/R = 1+2/2.
R = 2/3 Kohm.
V = i*R.
V = 3 mA*2/3 Kohm.
V = 2 v.
P = V*I.
P = 2v*3 mA.
P = 6 mW.
1/R = 1+2/2.
R = 2/3 Kohm.
V = i*R.
V = 3 mA*2/3 Kohm.
V = 2 v.
P = V*I.
P = 2v*3 mA.
P = 6 mW.
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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