Question

A coil in the heater consume power P on passing current. If it is cut into halves and joined in parallel, it will consume power:

• A. P
• B. 2P
• C. 4P
• D. 1/2P

Hint:

Cutting a wire into \(n\) equal parts and connecting them in parallel across the same voltage increases the power consumption by a factor of \(n^2\).
Here, \(n = 2\), so the power becomes \(2^2 P = 4P\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A heating coil with resistance \(R\) is connected across a constant voltage source \(V\), consuming power \(P\).
The coil is cut into two equal halves, and these halves are connected in parallel across the same source. We need to determine the new power consumption.

Step 2: Key Formula or Approach:
Power consumed is given by:
\[ P = \frac{V^2}{R_{eq}} \]
The resistance of a wire is directly proportional to its length: \(R \propto l\).

Step 3: Detailed Explanation:

• Let the initial resistance of the entire coil be \(R\).

• The initial power consumed is:
\[ P = \frac{V^2}{R} \]

• When the coil is cut into two equal halves, the length of each half becomes half of the original length. Since resistance is proportional to length, the resistance of each piece becomes:
\[ R_1 = R_2 = \frac{R}{2} \]

• These two halves are now connected in parallel. The equivalent resistance (\(R_{eq}\)) of this combination is:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R/2} + \frac{1}{R/2} = \frac{2}{R} + \frac{2}{R} = \frac{4}{R} \]
\[ R_{eq} = \frac{R}{4} \]

• The new power consumption \(P'\) across the same voltage \(V\) is:
\[ P' = \frac{V^2}{R_{eq}} = \frac{V^2}{R/4} = 4 \left( \frac{V^2}{R} \right) = 4P \]


Step 4: Final Answer:
The new power consumption of the parallel combination is \(4P\).