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

Options

A. `K_(1)+K_(2)`
B. `(K_(1)+K_(2))/(2)`
C. `(2K_(1)K_(2))/(K_(1)+K_(2))`
D. `(K_(1)+K_(2))/(2K_(1)K_(2))`

Answer

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

### Step-by-Step Solution:

1. **Understand the Configuration**:
We have two slabs made of different materials with thermal conductivities \( K_1 \) and \( K_2 \). Both slabs have the same thickness \( L \) and are arranged in parallel.

2. **Identify the Area**:
Let the cross-sectional area of each slab be \( A \). Since they are in parallel, the total area for heat conduction will be \( 2A \) (the area of both slabs combined).

3. **Calculate the Thermal Resistance**:
The thermal resistance \( R \) for a slab is given by the formula:
\[ R = \frac{L}{KA} \]
Therefore, for each slab:
- For the first slab (material with conductivity \( K_1 \)):
\[ R_1 = \frac{L}{K_1 A} \]
- For the second slab (material with conductivity \( K_2 \)):
\[ R_2 = \frac{L}{K_2 A} \]

4. **Use the Parallel Resistance Formula**:
When resistances are in parallel, the equivalent resistance \( R_{eq} \) is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
Substituting the values of \( R_1 \) and \( R_2 \):
\[ \frac{1}{R_{eq}} = \frac{K_1 A}{L} + \frac{K_2 A}{L} \]

5. **Combine the Terms**:
Factor out \( \frac{A}{L} \):
\[ \frac{1}{R_{eq}} = \frac{A}{L} \left( K_1 + K_2 \right) \]

6. **Find the Equivalent Conductivity**:
The equivalent thermal resistance can also be expressed in terms of the equivalent conductivity \( K_{eq} \):
\[ R_{eq} = \frac{L}{K_{eq} \cdot 2A} \]
Setting the two expressions for \( R_{eq} \) equal:
\[ \frac{L}{K_{eq} \cdot 2A} = \frac{1}{\frac{A}{L} (K_1 + K_2)} \]

7. **Solve for \( K_{eq} \)**:
Cross-multiplying gives:
\[ K_{eq} \cdot 2A = \frac{L^2}{A (K_1 + K_2)} \]
Rearranging gives:
\[ K_{eq} = \frac{K_1 + K_2}{2} \]

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