Question

When a potential difference V is applied across a wire of resistance R, it dissipates energy at a rate W. If the wire is cut into two halves and these halves are connected mutually parallel across the same supply, the same supply, the energy dissipation rate will become:

• A. $\frac{1}{4}W$
• B. $\frac{1}{2}W$
• C. 4W
• D. 2W

Answer 1

Answer: (C)

4W

Answer 2

To solve this problem, we need to analyze the energy dissipation rate before and after the wire is cut.

Initial Energy Dissipation Rate: The power P dissipated by a resistor is given by:

$P = \frac{V^2}{R}$,

where: - V is the voltage across the resistor, and R is the resistance.

Initially, with resistance R , the energy dissipation rate is:

$W = P = \frac{V^2}{R}$.

After Cutting the Wire: When the wire is cut into two halves, each half has a resistance of:

$R' = \frac{R}{2}$.

Connecting in Parallel: When these two halves are connected in parallel, the equivalent resistance R eq is given by:

$\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} = \frac{2}{R/2} = \frac{4}{R}$.

Thus, the equivalent resistance is:

$R_{eq} = \frac{R}{4}$.

New Energy Dissipation Rate: The new power P' dissipated in the circuit with the new resistance R eq is:

$P' = \frac{V^2}{R_{eq}} = \frac{V^2}{R/4} = \frac{4V^2}{R}$.

Comparing Power Dissipation Rates: Since $W = \frac{V^2}{R}$ from the original circuit, we can relate the new power:

$P' = 4W$.