We use Kirchhoff's current and voltage laws to solve for the unknown currents.
The first equation is given by:
\[ I_1 + I_3 - I_2 = -2 ag{1} \]
The second equation is:
\[ I_3 + 2I_2 = 5 ag{2} \]
The third equation is:
\[ 2I_2 - (I_3 - I_2) - (I_1 + I_3 - I_2) = 5 ag{3} \]
From equation (3):
\[ I_1 = -\frac{11}{5} \, \text{A} \]
Therefore:
\[ y = 11 \]
\(y = 11\)
The current passing through the battery in the given circuit, is:
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: