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.
Chandu said:
1 decade ago
PARALLEL CONNECTION:-
R eq =
[(2 X 1) (10^3 X 10^3)]/[(2+1) 10^3]
2/3 10^3 OHMS..........
P= I * I * R eq
(3*3*2/3)(10^-3)(10^-3)(10^3)
6mw ans
R eq =
[(2 X 1) (10^3 X 10^3)]/[(2+1) 10^3]
2/3 10^3 OHMS..........
P= I * I * R eq
(3*3*2/3)(10^-3)(10^-3)(10^3)
6mw ans
Naseem Ul Aziz said:
1 decade ago
p=vi
p1=?
p2=?
pt=?
p1=????
vt=ir
rt=666.666 ohm
vt=2 volt
p1=2m w
p2=4mw
pt=p1+p2=6 m w
p1=?
p2=?
pt=?
p1=????
vt=ir
rt=666.666 ohm
vt=2 volt
p1=2m w
p2=4mw
pt=p1+p2=6 m w
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
Syed said:
1 decade ago
I agree with ramprabhu answer.
Opion "A" is correct.
Opion "A" is correct.
Ramprabu said:
1 decade ago
P=I^2(R)
R=>2K AND 1K IN PARALLEL
HENCE EQUALENT R VALUE IS (2/3)
P=(3m)^2(2/3)
=>(9u)(2/3)
=>6uw
SO I CONCLUDE HERE OPTION A IS CORRECT
R=>2K AND 1K IN PARALLEL
HENCE EQUALENT R VALUE IS (2/3)
P=(3m)^2(2/3)
=>(9u)(2/3)
=>6uw
SO I CONCLUDE HERE OPTION A IS CORRECT
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
Sree said:
1 decade ago
The formula for resistor when 2 resistors are connected in parallel is 1/R=1/R1+1/R2.
Mamatha.m said:
1 decade ago
WKT,
P=I2*R
where Rp=R1R2/R1+R2
Rp=(2*1)/(2+1)
Rp=2/3k ohm
P=(3m)^2*(2/3)k
P=6mW
P=I2*R
where Rp=R1R2/R1+R2
Rp=(2*1)/(2+1)
Rp=2/3k ohm
P=(3m)^2*(2/3)k
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
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