Problem Statement: A 24 volt battery of internal resistance R=4 ohms is connected to a variable resistance r. The rate of heat dissipated in the resistor is Max when the current drawn from the battery is my. The current drawn from the battery will be I/2 when R is equal to?

Solution:
To solve this problem, we need to use the concepts of Ohm's law and power dissipation in resistors. Let's break down the problem into two parts:

Part 1: Maximum heat dissipation
When the rate of heat dissipation in the resistor is maximum, the power dissipated in the resistor is also maximum. The power dissipated in a resistor can be calculated using the formula:

P = I^2 * r

where P is the power dissipated, I is the current flowing through the resistor, and r is the resistance of the resistor.

Now, we know that the battery has a fixed voltage of 24 volts, and it is connected to a variable resistance r. So, the total resistance in the circuit is:

R_total = R + r

Using Ohm's law, we can calculate the current flowing through the circuit:

I = V / R_total

where V is the voltage of the battery.

Substituting the value of R_total, we get:

I = V / (R + r)

Now, we can substitute this value of I in the formula for power dissipation:

P = (V^2 / (R + r)^2) * r

To find the value of r that maximizes P, we need to take the derivative of P with respect to r and set it to zero:

dP/dr = (2V^2 / (R + r)^3) * r - (V^2 / (R + r)^2) = 0

Simplifying this equation, we get:

2r = R + r

r = R

Therefore, the resistance that maximizes the rate of heat dissipated in the resistor is equal to the internal resistance of the battery, which is R = 4 ohms.

Part 2: Current drawn from the battery
Now, we need to find the value of r that will result in a current of I/2 flowing through the circuit. Using Ohm's law, we can write:

I/2 = V / (R + r)

Solving for r, we get:

r = (2V / I) - R

Substituting the values of V and R, we get:

r = (2 * 24 / I) - 4

Substituting the value of I/2 for I, we get:

r = 40 - 2I

Therefore, the value of r that will result in a current of I/2 flowing through the circuit is given by the equation r = 40 - 2I.

Conclusion:
In summary, the rate of heat dissipated in the resistor is maximum when the resistance of the variable resistor r is equal to the internal resistance of the battery, which is R = 4 ohms. The current drawn from the battery will be I/2 when the resistance of the variable resistor r is equal to 40 - 2I.