in a 120-volt circuit having a resistance of 12 ohms, the power W in watts when a current I is flowing through is given by W=120 I - 12 I^2. what is the maximum power, in watts, that can be delivered in this circuit?
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w = 120 I - 12 i^2
To find the maximum value, we will need to find the derivativ'es zero.
==> w' = 120 - 24 I = 0
==> I = 120/24 = 5
Since the factor of I^2 is negative, then the function has Maximum value when I = 5
==> w = 120*I - 12 I^2
= 120*5 - 12*5^2
= 600 - 300 = 300
Then the maximum power is W = 300 watts.
The equation power w is given by :
W = 120i - 12i^2
To find the maximum power W.
To maximise W:
W/12 = (120i-12i^2)/12 = 10i-i^2
W/12 = 2*5i-i^2
W/12 =5^5 -5^2+2*5i-i^2
W/12 = 5^2 - (5 - i)^2
W/12 <5^2 , as (5-i)^2 being a perfect square is always positive.
So the W/12 is maximum and equal to 5^2=25 when i = 5 amp.
Or the maximum power that could drawn is W = 25*12 = 300 watts, when i = 5amp.
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