You are given resistacne wire of length 50 cm and a battery of negligible resistacne In which of the folowing cases is larges amount of heat generated?

Options

A. When the wire is connected to the battery directly
B. When the wire is divided into two parts and both the parts are connected to the battery oin parallel
C. When the wire is divided into four parts and all the four parts are connected to the battery in parallel
D. When only hjalf of the wire is connected to the battery

Explanation

To determine in which case the largest amount of heat is generated when a resistance wire is connected to a battery, we can use the formula for heat generated in a resistor, which is given by:

\[ H = \frac{V^2}{R} \times t \]

where:
- \( H \) is the heat generated,
- \( V \) is the voltage across the resistor,
- \( R \) is the resistance,
- \( t \) is the time for which the current flows.

Since the battery has negligible resistance, the voltage \( V \) remains constant in all cases. Therefore, the heat generated is inversely proportional to the resistance \( R \). This means that the smaller the resistance, the larger the heat generated.

Let's analyze each case step by step:

### Step 1: Case 1 - Wire connected directly to the battery
- The resistance of the wire is \( R_0 \).
- Heat generated, \( H_1 = \frac{V^2}{R_0} \).

### Step 2: Case 2 - Wire divided into two parts in parallel
- Each part has a resistance of \( \frac{R_0}{2} \).
- The equivalent resistance \( R_{eq} \) for two resistors in parallel is given by:
\[ R_{eq} = \frac{R/2 \times R/2}{R/2 + R/2} = \frac{R/4}{R/2} = \frac{R_0}{4} \]
- Heat generated, \( H_2 = \frac{V^2}{R_{eq}} = \frac{V^2}{R_0/4} = \frac{4V^2}{R_0} \).

### Step 3: Case 3 - Wire divided into four parts in parallel
- Each part has a resistance of \( \frac{R_0}{4} \).
- The equivalent resistance \( R_{eq} \) for four resistors in parallel is given by:
\[ R_{eq} = \frac{R/4 \times R/4 \times R/4 \times R/4}{R/4 + R/4 + R/4 + R/4} = \frac{R/16}{R} = \frac{R_0}{16} \]
- Heat generated, \( H_3 = \frac{V^2}{R_{eq}} = \frac{V^2}{R_0/16} = \frac{16V^2}{R_0} \).

### Step 4: Case 4 - Half of the wire connected to the battery
- The resistance of half the wire is \( \frac{R_0}{2} \).
- Heat generated, \( H_4 = \frac{V^2}{R_{eq}} = \frac{V^2}{R_0/2} = \frac{2V^2}{R_0} \).

### Step 5: Compare the heat generated in all cases
- \( H_1 = \frac{V^2}{R_0} \)
- \( H_2 = \frac{4V^2}{R_0} \)
- \( H_3 = \frac{16V^2}{R_0} \)
- \( H_4 = \frac{2V^2}{R_0} \)

From the calculations:
- \( H_3 \) (when the wire is divided into four parts and connected in parallel) is the largest.

### Conclusion
The case in which the largest amount of heat is generated is **Case 3**, where the wire is divided into four parts and all parts are connected to the battery in parallel.

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