Question 6A38

When a constant voltage is applied across a resistance, the power dissipated is inversely proportional to the resistance. This relationship is described by the formula $P = V^2/R$, where P is power in watts, V is voltage in volts, and R is resistance in ohms. If the resistance (R) is doubled to 2R while the voltage (V) remains constant, the new power dissipation will be: $P_{new} = V^2 / (2R)$ Comparing this to the original power ($P_{original} = V^2/R$), we see that: $P_{new} = (1/2) \times (V^2/R) = (1/2) \times P_{original}$ Therefore, doubling the resistance while keeping the voltage constant results in the power dissipation being reduced to one half of its original value.