If you halve the resistance in a circuit to which constant voltage is applied, what happens to the power dissipation?

When you halve the resistance in a circuit while keeping the voltage constant, the power dissipation in the circuit actually increases. This increase occurs due to the relationship defined by the formula for power in an electrical circuit, which is expressed as:

[ P = \frac{V^2}{R} ]

Where ( P ) is power, ( V ) is voltage, and ( R ) is resistance.

When you reduce the resistance ( R ) by half, the equation shows that power ( P ) becomes:

[ P' = \frac{V^2}{R/2} = \frac{V^2 * 2}{R} ]

This indicates that the new power dissipation ( P' ) is twice the original power ( P ). Thus, halving the resistance results in a doubling of power dissipation, which aligns with the answer provided.

Understanding this relationship is critical in circuit analysis, as it highlights how sensitive power dissipation is to changes in resistance, especially under constant voltage conditions.