Question

An electric coil is rated $400\text{ W}$, $200\text{ V}$. It is cut into two equal parts and connected in parallel to the same source of $200\text{ V}$. Calculate the percentage increase in energy produced per second.

• A. $100\%$
• B. $200\%$
• C. $300\%$
• D. $400\%$

Hint:

When a wire is cut into $n$ equal segments and rearranged in parallel, the total power output increases by a factor of $n^2$. Here, $n=2$, so power becomes $2^2 = 4$ times the original. A $4$-times increase means a value becomes $400\%$, which stands for a $300\%$ net increase!

Answer 1

The Correct Option is C

Understanding the Concept: The electrical resistance ($R$) of a uniform conductor wire depends directly on its length ($l$) via $R = \rho \frac{l}{A}$. Energy produced per second is exactly equal to the power dissipation ($P$), which can be calculated across a constant voltage supply $V$ using: \[ P = \frac{V^2}{R} \]
Step 1: Find the initial resistance and parameters.
Let the original coil have resistance $R$. The initial power output is: \[ P_{\text{initial}} = 400\text{ W} \quad \text{at} \quad V = 200\text{ V} \]
Step 2: Determine the new total parallel resistance.
When the coil is sliced into two identical halves, the length of each segment becomes $\frac{l}{2}$. Since resistance is directly proportional to length, each half gets a resistance of: \[ R' = \frac{R}{2} \] These two pieces are joined back in a parallel arrangement. The new equivalent resistance ($R_{\text{eq}}$) is: \[ \frac{1}{R_{\text{eq}} } = \frac{1}{R'} + \frac{1}{R'} = \frac{2}{R} + \frac{2}{R} = \frac{4}{R} \implies R_{\text{eq}} = \frac{R}{4} \]
Step 3: Calculate the final power dissipation and percentage increase.
The new power dissipation ($P_{\text{final}}$) connected to the original $200\text{ V}$ line is: \[ P_{\text{final}} = \frac{V^2}{R_{\text{eq}}} = \frac{V^2}{\left(\frac{R}{4}\right)} = 4 \left(\frac{V^2}{R}\right) = 4 \times P_{\text{initial}} \] \[ P_{\text{final}} = 4 \times 400\text{ W} = 1600\text{ W} \] The percentage increase in energy produced per second is: \[ % \text{ Increase} = \left(\frac{P_{\text{final}} - P_{\text{initial}}}{P_{\text{initial}}}\right) \times 100% \] \[ % \text{ Increase} = \left(\frac{1600 - 400}{400}\right) \times 100% = \frac{1200}{400} \times 100% = 300% \]