Two heating coils, one of fine wire and the other of thick wire of the same material and of the same length are connected in series and in parallel. Which of the following statement is correct ?

Options

A. In series fine wire liberates more energy while in parallel thick wire will liberate more energy.
B. In series fine wire liberates less energy while in parallel thick wire will liberate more energy
C. Both will liberate quickly
D. In series the thick wire will liberate more while in parallel it will liberate less energy

Answer

To solve the problem, we need to analyze the heating effect of two coils made of the same material and length but differing in thickness. We will consider their behavior when connected in series and in parallel.

### Step-by-Step Solution:

**Step 1: Determine the Resistance of Each Coil**
- The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.
- Since both coils are made of the same material and have the same length, the resistivity and length are constant. Therefore, the resistance is inversely proportional to the cross-sectional area \( A \).

**Step 2: Compare the Resistances of Fine and Thick Wires**
- Let \( R_1 \) be the resistance of the fine wire and \( R_2 \) be the resistance of the thick wire.
- Since the area of the fine wire \( A_1 \) is less than that of the thick wire \( A_2 \), we have:
\[ R_1 > R_2 \]
This means the fine wire has a greater resistance than the thick wire.

**Step 3: Analyze the Series Connection**
- When connected in series, the same current \( I \) flows through both resistors.
- The heat \( H \) liberated in a resistor can be calculated using the formula:
\[ H = I^2 R t \]
where \( t \) is the time.
- Since the current \( I \) and time \( t \) are the same for both resistors, the heat liberated will be directly proportional to their resistances:
\[ H_1 \propto R_1 \quad \text{and} \quad H_2 \propto R_2 \]
- Since \( R_1 > R_2 \), it follows that:
\[ H_1 > H_2 \]
Therefore, the fine wire liberates more energy when connected in series.

**Step 4: Analyze the Parallel Connection**
- When connected in parallel, the potential difference \( V \) across both resistors is the same.
- The heat liberated in this case can be calculated using:
\[ H = \frac{V^2}{R} t \]
- Since \( V \) and \( t \) are constant for both resistors, the heat liberated will be inversely proportional to their resistances:
\[ H_1 \propto \frac{1}{R_1} \quad \text{and} \quad H_2 \propto \frac{1}{R_2} \]
- Since \( R_1 > R_2 \), it follows that:
\[ H_1 < H_2 \]
Therefore, the thick wire liberates more energy when connected in parallel.

**Conclusion:**
- In series, the fine wire liberates more energy.
- In parallel, the thick wire liberates more energy.
- Thus, the correct statement is that "in series, the fine wire liberates more energy while in parallel, the thick wire will liberate more energy."

### Final Answer:
**Option 1 is correct.**

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