When an alternating current given by I = 10sin (120π)t passes through a 12Ω resistor, the power dissipated in the resistor is
1200W
600W
120W
30W
Explanation
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Discussions (25)
i disagree with the option A,,,
firstly, check for the definition of r.m.s
secondly,the formal for Ir.m.s =i0/√2
therefore I=10/√2
=7.071...
now,putting it into the formula for power =I²R
p=600
It's rms value in use for power not peak value.
Io=10
Irms=10/√2
P=I²R=50x12=600
you can't use peak current (|°) to solve convert to root mean square current first the answer is B
You people sha.
The current given was the peak current and it can’t be used for calculate the power dissipated.
So we have to convert it back to the Irms before we can use the current.
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The answer to this is supposed to be 600W.
Since this is AC, we are required to use the RMS value of current and not the instantaneous, this is represented as I_rms = (I_o)/sqrt(2)
Doing this, we have 10/sqrt(2), which is 7.07A
Putting it into the power formula: I^2 * R
(7.07A)^2 * 12
Which is equal to 600W ( 0.6kW )
Since Io is given,
Convert to rms.
So P=(I rms)²×R
P=(10/sqrt2)²×12
P=(10/1.414)²×12
P=600w
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