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

The Correct Option is C

To address this issue, we will evaluate the energy dissipation rate before and after the wire is severed.

Initial Energy Dissipation Rate: The power P expended by a resistor is defined by:

\( P = \frac{V^2}{R} \),

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

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

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

After Wire Severing: Upon bisecting the wire, each resulting segment possesses a resistance of:

\( R' = \frac{R}{2} \).

Parallel Connection: When these two segments are joined in parallel, the aggregate resistance R_eq is determined by:

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

Consequently, the equivalent resistance is:

\( R_{eq} = \frac{R}{4} \).

New Energy Dissipation Rate: The revised power P' dissipated within the circuit, given the new resistance R_eq, is:

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

Comparative Power Dissipation: As \( W = \frac{V^2}{R} \) was established for the original circuit, the new power can be expressed as:

\( P' = 4W \).