The power dissipated in a resistor is given by \( P = I^2 R \). If the current is doubled, i.e., \( I \rightarrow 2I \), the power becomes: \[ P' = (2I)^2 R = 4I^2 R. \]
Thus, the power quadruples when the current is doubled.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: