Electronics - Ohm's Law - Discussion
Discussion Forum : Ohm's Law - General Questions (Q.No. 11)
11.
What is the power in the given circuit?
Discussion:
21 comments Page 2 of 3.
Mandala suresh said:
1 decade ago
P = V*I.
We know V = I*R.
So P = I^2*R.
P = 400*10^-3*400*10^-3*200.
P = 32 W.
We know V = I*R.
So P = I^2*R.
P = 400*10^-3*400*10^-3*200.
P = 32 W.
Chandu said:
1 decade ago
p=I^2R
AMAR PRATAP said:
1 decade ago
P = V*I.
V = I*R.
P = I^2*R.
P = 400*10^-3*400*10^-3*200.
P = 32000000*10^-6.
P = 32W RIGHT ANSWER.
V = I*R.
P = I^2*R.
P = 400*10^-3*400*10^-3*200.
P = 32000000*10^-6.
P = 32W RIGHT ANSWER.
Rathi said:
1 decade ago
@Praveen.
Your explanation wrong.
400 x 10^-3*400 x 10^-3*200 = 32W.
Your explanation wrong.
400 x 10^-3*400 x 10^-3*200 = 32W.
Badar said:
1 decade ago
p=I*I*R
p=(400/1000)*(400/1000)*200
p=(1600/10000)*200
p=(16*200)/100
p=3200/100
p=32W
p=(400/1000)*(400/1000)*200
p=(1600/10000)*200
p=(16*200)/100
p=3200/100
p=32W
Karthik said:
1 decade ago
V=IR
P=VI
P=I*I*R
I=400/1000=0.4A
I=0.4*0.4=0.16A
P=0.16*200=32W
P=VI
P=I*I*R
I=400/1000=0.4A
I=0.4*0.4=0.16A
P=0.16*200=32W
Mainpal singh said:
1 decade ago
V=IR
P=VI
P=I*I*R
I=400/1000=0.4A
I=0.4*0.4=0.16A
P=0.16*200=32W
P=VI
P=I*I*R
I=400/1000=0.4A
I=0.4*0.4=0.16A
P=0.16*200=32W
Praveen said:
1 decade ago
Given ans is wrong 32kw is the correct ans
p=i^2*r
400*400*200*10^-3
=32000=32kw
p=i^2*r
400*400*200*10^-3
=32000=32kw
MD.Abdul Qualique said:
1 decade ago
p=v*I
but we know from ohm's law that
V=IR
so,
p=I^2*R
here we also need to see the units
solution: i am explaining it by elaborating
P=400x10^-3x400x10^-3x200
P=32000000 x 10^-6
P=32x10^6x10^-6
ANS:P=32w
but we know from ohm's law that
V=IR
so,
p=I^2*R
here we also need to see the units
solution: i am explaining it by elaborating
P=400x10^-3x400x10^-3x200
P=32000000 x 10^-6
P=32x10^6x10^-6
ANS:P=32w
Dilipkumar M T said:
1 decade ago
This is wrong
V=I*R
V=I*R
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