Electronics - Ohm's Law - Discussion

Discussion Forum : Ohm's Law - General Questions (Q.No. 11)
11.

What is the power in the given circuit?

32 W
80 W
500 W
16 kW
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
21 comments Page 2 of 3.

DANTHEMAN said:   1 decade ago
Lets make this easy!

400mA is equal to 400/1000 = 0.4 Amps (moving the decimal place three places to the left gives you milli).

0.4 x 0.4 = 0.16.

0.16 x 200 = 32w.

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.

Rathi said:   1 decade ago
@Praveen.

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

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

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

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

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

Dilipkumar M T said:   1 decade ago
This is wrong
V=I*R

Preethi said:   1 decade ago
Not able to understand this.


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