Question

the formula for the power output P of a battery is P=V*I-R*I^2, where V is the electromotive force in volts, R is the resistance in ohms, and I is the current in amperes. Find the current that corresponds to a maximum value of P in a battery for which V = 12 volts and R =0.5 ohm. Assume that a 15-ampere fuse bounds the output in the interval 0 ≤ I ≤ 15.

Answer 1

Formula for power

P=VI-RI²

Here,V is the voltage, I is the current, and R is the resistance.

Substitute V = 12 V  and R = 0.5 ohm.

P =12I - 0.5I²   0 ≤ I ≤ 15.

Take derivative both sides with respect to I

dP/dI  = 12 - I

For maximum power dP/dI  =0

12 - I = 0

I  =  12 amps

Substitute I  =  12 amps in power equation.

P_max=12(12) - 0.5(12)²

         = 72 W

No,the 15 amp fuse not increases , yhe power output decreases for I >12  .