Two metal plates of same area and thickness `l_(1)` and `l_(2)` are arranged in series If the thermal conductivities of the materials of the two plates are `K_(1)` and `K_(2)` The thermal conductivity of the combination is .

Options

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

Explanation

To find the effective thermal conductivity of two metal plates arranged in series, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
We have two plates with the same area (A) and different thicknesses (L1 and L2). The thermal conductivities of the materials are K1 and K2, respectively.

2. **Define Thermal Resistance**:
The thermal resistance (R) for each plate can be defined using the formula:
\[ R = \frac{L}{K \cdot A} \]
where L is the thickness of the plate, K is the thermal conductivity, and A is the area.

3. **Calculate Thermal Resistance for Each Plate**:
- For Plate 1 (thickness L1, conductivity K1):
\[ R_1 = \frac{L_1}{K_1 \cdot A} \]
- For Plate 2 (thickness L2, conductivity K2):
\[ R_2 = \frac{L_2}{K_2 \cdot A} \]

4. **Total Thermal Resistance**:
Since the plates are in series, the total thermal resistance (R_effective) is the sum of the individual resistances:
\[ R_{\text{effective}} = R_1 + R_2 = \frac{L_1}{K_1 \cdot A} + \frac{L_2}{K_2 \cdot A} \]

5. **Combine the Terms**:
Factor out the area (A) from the equation:
\[ R_{\text{effective}} = \frac{1}{A} \left( \frac{L_1}{K_1} + \frac{L_2}{K_2} \right) \]

6. **Relate to Effective Thermal Conductivity**:
The effective thermal conductivity (K_effective) can be defined as:
\[ R_{\text{effective}} = \frac{L_{\text{total}}}{K_{\text{effective}} \cdot A} \]
where \( L_{\text{total}} = L_1 + L_2 \).

7. **Set the Equations Equal**:
Equating the two expressions for R_effective gives:
\[ \frac{L_1 + L_2}{K_{\text{effective}} \cdot A} = \frac{1}{A} \left( \frac{L_1}{K_1} + \frac{L_2}{K_2} \right) \]

8. **Solve for K_effective**:
Cancel A from both sides and rearrange:
\[ K_{\text{effective}} = \frac{(L_1 + L_2) \cdot (K_1 \cdot K_2)}{L_1 \cdot K_2 + L_2 \cdot K_1} \]

### Final Answer:
The effective thermal conductivity of the combination is:
\[ K_{\text{effective}} = \frac{(L_1 + L_2) \cdot (K_1 \cdot K_2)}{L_1 \cdot K_2 + L_2 \cdot K_1} \]

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