A 24 V battery of internal resistance `4.0 Omega` is connected to a variable resistance. At what value of the current drawn from the battery is the rate of heat produced in the resistor maximum?

Options

A. (a)2A
B. (b)3A
C. (c)4A
D. (d)6A

Answer

To solve the problem step by step, we need to determine the current at which the rate of heat produced in the resistor is maximum.

### Step 1: Understand the circuit
We have a battery with an electromotive force (emf) of \( E = 24 \, V \) and an internal resistance \( r = 4 \, \Omega \). The battery is connected to a variable resistance \( R \).

### Step 2: Condition for maximum heat
The rate of heat produced in the resistor is maximum when the resistance of the resistor \( R \) is equal to the internal resistance \( r \) of the battery. Therefore, we set:
\[ R = r = 4 \, \Omega \]

### Step 3: Calculate total resistance
The total resistance \( R_{total} \) in the circuit is the sum of the internal resistance and the variable resistance:
\[ R_{total} = r + R = 4 \, \Omega + 4 \, \Omega = 8 \, \Omega \]

### Step 4: Calculate the current
Using Ohm's law, the current \( I \) drawn from the battery can be calculated as:
\[ I = \frac{E}{R_{total}} = \frac{24 \, V}{8 \, \Omega} = 3 \, A \]

### Conclusion
The current drawn from the battery at which the rate of heat produced in the resistor is maximum is:
\[ \boxed{3 \, A} \]