A 24 V battery of internal resistance 4 ohm is connected to a variable resistor. At what value of the current drawn from the battery is the rate of heat produced in the resistor is maximum?

Answer 1

let P be the power delivered to extenal resistance R.

P= I^2*R= V^2/(4+R)^2 * R
now will calculus, dP/dR=0 or
0=V^2/(4+R)^2 + -2RV^2/(4+R)
0=1-2R(4+R)
2R^2-8R-1=0
(R-4)(2R+.25)=0
R=4 ohms, or R= -1/4 ohms. In real resistsors, resistance can be negative.

So max current= 24/8=3 amps at max heating of R