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 1 of 4.
Moksha said:
7 years ago
p = I^2R.
1/R=(1+2)/(1*2) = 3/2,
R = 2/3k ohms,
p = (3m)^2 * 2/3k = 6mw.
1/R=(1+2)/(1*2) = 3/2,
R = 2/3k ohms,
p = (3m)^2 * 2/3k = 6mw.
(1)
Rashmi said:
1 decade ago
V = IR BY OHM'S LAW.
From the following data:
I = 3 m.
R1 = 1k.
R2 = 2k.
When resistors are connected in parallel.
Req is given by Req = R1*R2/R1+R2.
P = I^2*Req.
P = (3m)^2*(2/3)k.
P = 6 mW.
From the following data:
I = 3 m.
R1 = 1k.
R2 = 2k.
When resistors are connected in parallel.
Req is given by Req = R1*R2/R1+R2.
P = I^2*Req.
P = (3m)^2*(2/3)k.
P = 6 mW.
(1)
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)
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.
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.
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.
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
Abc said:
1 decade ago
What would be the total power loss had the circuit been connected in series?
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.
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.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers