Here, \(R_2\), \(R_3\), and \(R_4\) are in parallel. The equivalent resistance of these resistors is given by:
\[ \frac{1}{R_{234}} = \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} = \frac{1}{8} + \frac{1}{4} + \frac{1}{8} \]
\[ R_{234} = 2 \, \Omega \]
This resistance is in series with \(R_1\), so the total equivalent resistance is:
\[ R_{\text{total}} = R_1 + R_{234} = 10 \, \Omega + 2 \, \Omega = 12 \, \Omega \]
The current supplied by the battery is:
\[ I = \frac{V}{R_{\text{total}}} = \frac{12 \, V}{12 \, \Omega} = 1 \, \text{A} \]
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: