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 1 of 3.
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
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.
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)
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.
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.
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.
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)
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.
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.
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.
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.
V+ IxR 0.4x200 = 80 volts.
Power = VxI.
80x0.4 = 32watts.
(1)
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.
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
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.
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