FCC Exam Question: 6A38
A resistance across which a constant voltage is applied is doubled. What power dissipation will result?
Explanation: 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.
6A194
6A282
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Includes Elements 1, 3, 6, 7R, 8, and 9.