Ohm's law: current against voltage

Three resistors, of $8\ \Omega$, $20\ \Omega$ and $40\ \Omega$, are each joined to voltages from $0$ to $12\ \text{V}$. You will make the program find the current through each at $12\ \text{V}$, the resistance of an unknown part from two readings, and the power in the $20\ \Omega$ resistor at $9\ \text{V}$. Then it draws the current against the voltage.

The idea. $I = \dfrac{V}{R}$, so on a graph of $I$ against $V$ each resistor is a straight line with slope $\dfrac{1}{R}$. Power is $P = \dfrac{V^2}{R}$.

Your tasks.

1. Inside the loop, work out I12, the current at $12\ \text{V}$ through the resistor R.
2. Find R_unknown from the two readings $(2\ \text{V}, 0.16\ \text{A})$ and $(10\ \text{V}, 0.8\ \text{A})$.
3. Find P9, the power in the $20\ \Omega$ resistor at $9\ \text{V}$.