The power supply is switched on for 300s and Calculate (i) the charge, Q, that flows in the coil. (ii) electrical resistance, R, of the coil.

Step 1: Identify the given values from the problem and the student's work. The power supply is switched on for $t = 300s$. From the student's work in part (i), the current $I = 2.5A$ is used. We will assume this is the current flowing in the coil.

Step 2: Calculate the charge, Q, that flows in the coil. The formula relating current (I), charge (Q), and time (t) is $I = \frac{Q}{t}$. To find the charge, we rearrange the formula to $Q = I \times t$. $Q = 2.5A \times 300s$ $Q = 750C$ The charge, Q, that flows in the coil is $\boxed{750 C}$.

Step 3: Calculate the electrical resistance, R, of the coil. The problem does not explicitly state the components of the coil. However, the student's work shows $R = R_1 + R_2 = 2.5 \Omega + 2.5 \Omega$. We will assume that the coil consists of two resistors, each with a resistance of $2.5 \Omega$, connected in series. For resistors in series, the total resistance is the sum of individual resistances. $R = R_1 + R_2$ $R = 2.5 \Omega + 2.5 \Omega$ $R = 5 \Omega$ The electrical resistance, R, of the coil is $\boxed{5 \Omega}$.