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 1 of 3.

Jayashri nb said:   3 years ago
P = I^2 * R.
P = 400 * 10^-3 * 400 * 10^-3 * 200.
= 32000000 * 10^-6,
= 32 * 10^6 * 10^-6,
Cancel 10^6 and 10^-6,
Ans 32.
Then P = 32watt.
(1)

Shraddha said:   7 years ago
P=v^2*R.
V=(400/1000)*200,
V=80,
P=v^2/R.
P=(80)^2/200,
ANS: P=32w.

Sahal waf said:   7 years 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.
P=400x10^-3x400x10^-3x200.
P=32000000 x 10^-6.
P=32x10^6x10^-6.
Answer : P = 32w.
(1)

M. V. NAGARAJU said:   7 years ago
Thank you all for explaining the solution.

Tarun said:   8 years ago
V= R*I
V=200*400
V=8000V.

then,
P=V*I.
If we have found all values of in this equation,
P=8000*400,
P=3200000 or 32 *10 sq5.

Omprakash said:   8 years ago
V=RI
V=400*200/10000
V=8.

P=VI
P=8*4
P=32w.

Aditya said:   9 years ago
Given I = 400mA = 0.4A.

R = 200 ohms.

As we know that,
P = I^2 * R.

So, P = 0.4 * 0.4 * 200.
P = 32Volts.
(1)

Jose said:   1 decade ago
I = 400ma = 400/1000 = 0.4amps and R = 200 ohms.
V+ IxR 0.4x200 = 80 volts.

Power = VxI.
80x0.4 = 32watts.
(1)

Shawn said:   1 decade ago
What is p?

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.


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