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Power generated across a uniform wire co...

Power generated across a uniform wire connected across a supply is `H`. If the wire is cut into `n` equal parts and all the parts are connected in parallel across the same supply, the total power generated in the wire is

A

`(H)/(n^(2))`

B

`n^(2)H`

C

`nh`

D

`(H)/(n)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the power generated in a uniform wire when it is cut into `n` equal parts and connected in parallel across the same supply. ### Step-by-Step Solution: 1. **Initial Power Calculation**: - Let the original wire have a resistance \( R \) and a voltage \( V \) applied across it. - The power generated in the wire is given by the formula: \[ H = \frac{V^2}{R} \] 2. **Cutting the Wire**: - When the wire is cut into `n` equal parts, the length of each part is \( \frac{L}{n} \) (where \( L \) is the original length of the wire). - Since resistance \( R \) is directly proportional to length, the resistance of each part \( R' \) will be: \[ R' = \frac{R}{n} \] 3. **Connecting in Parallel**: - When these `n` parts are connected in parallel, the total resistance \( R_{total} \) of the parallel combination can be calculated using the formula for resistors in parallel: \[ \frac{1}{R_{total}} = \frac{1}{R'} + \frac{1}{R'} + \ldots + \frac{1}{R'} \quad (n \text{ times}) \] - This simplifies to: \[ \frac{1}{R_{total}} = n \cdot \frac{1}{R'} = n \cdot \frac{n}{R} \implies R_{total} = \frac{R}{n^2} \] 4. **Power in the New Configuration**: - The voltage across the parallel combination remains \( V \). - The new power \( H' \) generated in this configuration is given by: \[ H' = \frac{V^2}{R_{total}} = \frac{V^2}{\frac{R}{n^2}} = \frac{V^2 \cdot n^2}{R} \] - We know from the initial case that \( H = \frac{V^2}{R} \). Therefore, we can express \( H' \) in terms of \( H \): \[ H' = n^2 \cdot H \] 5. **Conclusion**: - The total power generated in the wire when cut into `n` equal parts and connected in parallel is: \[ H' = n^2 H \] ### Final Answer: The total power generated in the wire when cut into `n` equal parts and connected in parallel is \( n^2 H \).
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