A slab consists of two layers of different materials of the same thickness and having thermal conductivities `K_(1)` and `K_(2)`. The equivalent thermal conductivity of the slab is

Options

A. K1 + K2
B. K1 + K2/2
C. 2K1 + K2 /K1 + k2
D. K1 + K2/2KK2

Explanation

To find the equivalent thermal conductivity of a slab consisting of two layers of different materials with the same thickness, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Let the thermal conductivities of the two materials be \( K_1 \) and \( K_2 \).
- Let the thickness of each layer be \( L \).
- Let the cross-sectional area of each layer be \( A \).

2. **Determine the Thermal Resistances**:
- The thermal resistance \( R_1 \) for the first layer is given by:
\[ R_1 = \frac{L}{K_1 \cdot A} \]
- The thermal resistance \( R_2 \) for the second layer is given by:
\[ R_2 = \frac{L}{K_2 \cdot A} \]

3. **Calculate the Equivalent Resistance**:
- Since the two layers are in parallel, the equivalent thermal resistance \( R_{eq} \) can be calculated using the formula:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
- Substituting the expressions for \( R_1 \) and \( R_2 \):
\[ \frac{1}{R_{eq}} = \frac{K_1 \cdot A}{L} + \frac{K_2 \cdot A}{L} \]

4. **Simplify the Equation**:
- Factor out common terms:
\[ \frac{1}{R_{eq}} = \frac{A}{L} (K_1 + K_2) \]
- Therefore, we can express \( R_{eq} \) as:
\[ R_{eq} = \frac{L}{A(K_1 + K_2)} \]

5. **Relate Equivalent Resistance to Equivalent Conductivity**:
- The equivalent thermal resistance can also be expressed in terms of the equivalent thermal conductivity \( K_{eq} \):
\[ R_{eq} = \frac{L}{K_{eq} \cdot (2A)} \]
- Here, \( 2A \) is used because the total area for the parallel combination is twice the area of one layer.

6. **Set the Two Expressions for \( R_{eq} \) Equal**:
- Equating the two expressions for \( R_{eq} \):
\[ \frac{L}{A(K_1 + K_2)} = \frac{L}{K_{eq} \cdot (2A)} \]

7. **Solve for \( K_{eq} \)**:
- Cancel \( L \) and \( A \) from both sides:
\[ \frac{1}{K_1 + K_2} = \frac{1}{2K_{eq}} \]
- Rearranging gives:
\[ 2K_{eq} = K_1 + K_2 \]
- Thus, the equivalent thermal conductivity is:
\[ K_{eq} = \frac{K_1 + K_2}{2} \]

### Final Answer:
The equivalent thermal conductivity of the slab is:
\[ K_{eq} = \frac{K_1 + K_2}{2} \]