In a 120-circuit having resistance of 12 ohms, the power W in watts when a current I ( in amperes) is flowing is given by W=120I - 12I². Determine the maximum power that can be delivered in this circuit.
2007-01-27 09:19:14 · 3 answers · asked by ₪ڠYiffniff ڠ₪ 5 in Science & Mathematics ➔ Mathematics
Since W = 120I - 12I^2
At a turning point either maximum or minimum the gradient of the given expression must equal zero.
The gradient of the given expression can be gotten by differentiating the expression like so,
0 = 120 - 24I
so that 120 = 24I.
Thus I = 120/24
=5amps.
Observe that the given expression can only possess a maximum point.
Putting the value of I into the original expression, we have
120(5) - 12(5)^2
= 600 -300W
= 300W
2007-01-27 09:33:10 · answer #1 · answered by agboola f 2 · 1⤊ 0⤋
In a circuit with source impedance of 12 ohms, the maximum power is delivered when the load is also 12 ohms. The current to the load will be 120/24, and the voltage on the load will be (1/2)*120 = 60v. The maximum power will then be 60*120/24= 300W
2007-01-27 17:30:25 · answer #2 · answered by gp4rts 7 · 1⤊ 0⤋
This is a reverse parabola, thus you need to find its base. The easiest way is to take the derivative of W, which is W'(I) = 120-24I, thus 0 = 120-24I is where W reaches slope 0 (at its base). We get that I = 5, and thus the maximum power is W = 120(5)-12(5)^2=300.
2007-01-27 17:28:18 · answer #3 · answered by goldenflaws 2 · 1⤊ 0⤋