Two metallic wires of the same material and same length have different diameters. If we connect them in series across a battery, the heat produced is `H _(1).` If we connect them in parallel to the same battery the heat produced during the same time is `H_(2).` From the above, we infer that

Options

A. `H_(1) gt H _(2)`
B. `H_(1) lt H _(2)`
C. `H _(1) = H _(2)`
D. nothing can be decided

Explanation

To solve the problem, we need to analyze the heat produced in two metallic wires connected in series and parallel. Let's denote the two wires as Wire 1 and Wire 2, with different diameters but the same material and length.

### Step-by-Step Solution:

1. **Identify the Resistance of Each Wire:**
- The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
- Where \( \rho \) is the resistivity of the material, \( L \) is the length, and \( A \) is the cross-sectional area.
- For wires of the same length and material, the resistance will depend on the area of cross-section, which is related to the diameter \( d \):
\[ A = \frac{\pi d^2}{4} \]
- Therefore, if Wire 1 has diameter \( d_1 \) and Wire 2 has diameter \( d_2 \), we can express their resistances as:
\[ R_1 = \frac{\rho L}{A_1} = \frac{4\rho L}{\pi d_1^2} \]
\[ R_2 = \frac{\rho L}{A_2} = \frac{4\rho L}{\pi d_2^2} \]
- Since \( d_1 < d_2 \) (assuming Wire 1 is thinner), we have \( R_1 > R_2 \).

2. **Heat Produced in Series Connection:**
- When connected in series, the total resistance \( R_s \) is:
\[ R_s = R_1 + R_2 \]
- The heat produced \( H_1 \) in a time \( t \) when connected to a voltage \( V \) is given by:
\[ H_1 = \frac{V^2 t}{R_s} = \frac{V^2 t}{R_1 + R_2} \]

3. **Heat Produced in Parallel Connection:**
- When connected in parallel, the total resistance \( R_p \) is given by:
\[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \implies R_p = \frac{R_1 R_2}{R_1 + R_2} \]
- The heat produced \( H_2 \) in the same time \( t \) is:
\[ H_2 = \frac{V^2 t}{R_p} = \frac{V^2 t (R_1 + R_2)}{R_1 R_2} \]

4. **Comparing \( H_1 \) and \( H_2 \):**
- Now, we compare \( H_1 \) and \( H_2 \):
\[ \frac{H_1}{H_2} = \frac{\frac{V^2 t}{R_1 + R_2}}{\frac{V^2 t (R_1 + R_2)}{R_1 R_2}} = \frac{R_1 R_2}{(R_1 + R_2)^2} \]
- Since \( R_1 > R_2 \), it follows that \( R_1 + R_2 > R_1 \) and \( R_1 + R_2 > R_2 \). Therefore, the term \( R_1 + R_2 \) squared will always be greater than \( R_1 R_2 \):
\[ H_2 > H_1 \]

### Conclusion:
From the analysis, we conclude that:
\[ H_2 > H_1 \]