The total heat produced in resistor `r` in an `RL` circuit when the current in the inductor decreases from `I_(o)` to `0` is

Options

A. `LI_(0)^(2)`
B. `(1)/(2) LI_(0)^(2)`
C. `(3)/(2) LI_(0)^(2)`
D. `(1)/(3) LI_(0)^(2)`

Explanation

To find the total heat produced in a resistor \( R \) in an \( RL \) circuit when the current in the inductor decreases from \( I_0 \) to \( 0 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Power Dissipation**:
The power \( P \) dissipated in the resistor \( R \) is given by the formula:
\[ P = I^2 R \]
where \( I \) is the current flowing through the resistor.

2. **Current Decay in the RL Circuit**:
In an \( RL \) circuit, the current \( I(t) \) at time \( t \) when the switch is opened (or when the current is decreasing) is given by:
\[ I(t) = I_0 e^{-\frac{R}{L}t} \]
where \( I_0 \) is the initial current, \( R \) is the resistance, and \( L \) is the inductance.

3. **Total Energy (Heat) Calculation**:
The total energy \( W \) dissipated in the resistor as heat when the current changes from \( I_0 \) to \( 0 \) can be found by integrating the power over time:
\[ W = \int_0^\infty P \, dt = \int_0^\infty I^2 R \, dt \]
Substituting \( I(t) \):
\[ W = \int_0^\infty (I_0 e^{-\frac{R}{L}t})^2 R \, dt \]
\[ W = R I_0^2 \int_0^\infty e^{-\frac{2R}{L}t} \, dt \]

4. **Evaluating the Integral**:
The integral \( \int_0^\infty e^{-\frac{2R}{L}t} \, dt \) can be evaluated as follows:
\[ \int_0^\infty e^{-kt} \, dt = \frac{1}{k} \quad \text{(where \( k = \frac{2R}{L} \))} \]
Thus,
\[ \int_0^\infty e^{-\frac{2R}{L}t} \, dt = \frac{L}{2R} \]

5. **Substituting Back**:
Now substituting this result back into the expression for \( W \):
\[ W = R I_0^2 \cdot \frac{L}{2R} \]
The \( R \) cancels out:
\[ W = \frac{1}{2} I_0^2 L \]

6. **Final Result**:
Therefore, the total heat produced in the resistor \( R \) when the current decreases from \( I_0 \) to \( 0 \) is:
\[ W = \frac{1}{2} L I_0^2 \]

### Final Answer:
The total heat produced in the resistor \( R \) is \( \frac{1}{2} L I_0^2 \).