Electronics - Parallel Circuits - Discussion

4. 

What is the total power loss if 2 k and 1 k parallel-connected resistors have an IT of 3 mA?

[A]. 6 W
[B]. 36 W
[C]. 6 mW
[D]. 36 mW

Answer: Option C

Explanation:

No answer description available for this question.

Rama said: (Oct 31, 2010)  
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

Mamatha.M said: (Feb 14, 2011)  
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

Sree said: (Feb 17, 2011)  
The formula for resistor when 2 resistors are connected in parallel is 1/R=1/R1+1/R2.

Gautam said: (Mar 1, 2011)  
Here r=2000/3
& i^2=(3*10^-3)^2=9*10^-6
we know P=i^2*r=6*10^-3=6mW

Ramprabu said: (Mar 26, 2011)  
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

Syed said: (Apr 10, 2011)  
I agree with ramprabhu answer.

Opion "A" is correct.

Devaki said: (Apr 28, 2011)  
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

Naseem Ul Aziz said: (May 6, 2011)  
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

Chandu said: (May 16, 2011)  
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

Ramakant said: (Jul 7, 2011)  
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

Gopi said: (Sep 22, 2011)  
P=I^2*R
R=(2/3)*10^-3
I=3*10^-3
P=(3*10^-3)(3*10^-3)(2/3)*10^-3
P=6w

Nandini said: (Oct 30, 2011)  
Yes, I also agree vit rambabu.

Sunil Kumar said: (Nov 8, 2011)  
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

Vinay said: (Nov 16, 2011)  
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

Bk Rout said: (Nov 20, 2011)  
Vinays answer is right. So option C is correct.

Piyush Suthar said: (Dec 14, 2011)  
Thats simple 1/r=1/2+1/1
r=2/3
p=i^2*r
=(3ma)^2*2/3
=9ma*2/3
=6mw

Linton said: (Jan 13, 2012)  
@piyush suthar

How 1/2 + 1/1 will makes 2/3 ?

1/2 = (.5)+(1) = 1.5 = 3/2 not 2/3.

Abhishek Gupta said: (Feb 5, 2012)  
1*2=1/2*3=2*3=6W

Jonathan said: (Apr 8, 2012)  
Ramprabu is wrong, he should include 10^3 in total resistance, you'll get ((2/3)x10^3)(3mA)^2 = 6mW

Anjali.R said: (Apr 10, 2012)  
P=i^2*r
P=3^2*(2/3)
P=6mw

Sudhakar said: (Jun 15, 2012)  
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.

U.Suruthi said: (Jul 30, 2012)  
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.

Prasanna said: (Aug 19, 2012)  
Ramprabhu is wrong,
Resistance value here given is Kohms and he consider it to be only ohms

Abc said: (Oct 31, 2012)  
What would be the total power loss had the circuit been connected in series?

Vinod said: (Dec 5, 2013)  
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.

Hims said: (Mar 1, 2014)  
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.

Irshad K A said: (May 3, 2014)  
R = 2/2+1/2.
R = 3/2.
R = 2/3.
R = 0.6668.

P = I2R.
P = 3*3*0.6668.
P = 6(approximately).

Asha said: (Oct 11, 2014)  
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.

Azhagusurya said: (Feb 20, 2015)  
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.

Rashmi said: (Jun 24, 2015)  
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.

Ana said: (Nov 14, 2015)  
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.

Addy said: (Jul 29, 2016)  
Simply,

Low resistance = Higher current flow.
So, we get applied voltage by dividing current in the ratio
2k/1k = 3mA.
1k = 2mA and 2k = 1mA.
Hence, applied voltage is 2V.
Now, we have I = 3mA, V = 2v.
P = V * I.
= 2 * 3mA.
= 6mA.

ANSWER 'C' IS CORRECT.

Chandrashekhar Manekar said: (Aug 25, 2016)  
Actually power( P) = I^2(R).

By using this, we can solve the problem.

Krish said: (Sep 10, 2016)  
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.

Divya said: (Dec 13, 2016)  
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.

Pijush Patra said: (Jul 7, 2017)  
What is the total resistance of 5 kilo ohm, 5mege ohm connected in parallel?

Shah Qaisar said: (Nov 10, 2018)  
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.

Sameer said: (Jan 8, 2019)  
Power=Voltage * Current.
Voltage= Current * Resistance,
Resistors in parallel (1/R=1/R1+1/R2) OR (R=R1*R2/R1+R2),
so, 1/R= 1/2+1/1 = 3/2KOhm , R= 2/3 KOhm,
P=V*I = (I*R)* I.
P= I*I*R,
P=3*3*2/3,
P=9*2/3,
P=18/3,
P=6mW.

Moksha said: (Feb 1, 2019)  
p = I^2R.
1/R=(1+2)/(1*2) = 3/2,
R = 2/3k ohms,
p = (3m)^2 * 2/3k = 6mw.

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