1. **Problem Statement:** We have an alternating current measured at equal intervals of 5 ms with values: Time (ms) = 0, 5, 10, 15, 20, 25, 30 and Current (A) = 0, 0.9, 2.6, 4.9, 5.8, 3.5, 0. We need to plot current against time, estimate the area under the curve over 30 ms using the mid-ordinate rule, and determine the mean current value. 2. **Formula and Explanation:** The mid-ordinate rule estimates the area under a curve by summing the values of the function at midpoints of intervals multiplied by the interval width. Area $\approx \sum f(m_i) \times \Delta t$ where $m_i$ are midpoints. Mean value of current $= \frac{\text{Area under curve}}{\text{Total time}}$. 3. **Calculate midpoints and their current values:** Intervals: 0-5, 5-10, 10-15, 15-20, 20-25, 25-30 ms Midpoints: 2.5, 7.5, 12.5, 17.5, 22.5, 27.5 ms We approximate current at midpoints by averaging current at endpoints: - At 2.5 ms: $\frac{0 + 0.9}{2} = 0.45$ A - At 7.5 ms: $\frac{0.9 + 2.6}{2} = 1.75$ A - At 12.5 ms: $\frac{2.6 + 4.9}{2} = 3.75$ A - At 17.5 ms: $\frac{4.9 + 5.8}{2} = 5.35$ A - At 22.5 ms: $\frac{5.8 + 3.5}{2} = 4.65$ A - At 27.5 ms: $\frac{3.5 + 0}{2} = 1.75$ A 4. **Calculate area using mid-ordinate rule:** Interval width $\Delta t = 5$ ms $$\text{Area} = 5 \times (0.45 + 1.75 + 3.75 + 5.35 + 4.65 + 1.75)$$ $$= 5 \times 17.7 = 88.5 \text{ A·ms}$$ 5. **Calculate mean current:** Total time = 30 ms $$\text{Mean current} = \frac{88.5}{30} = 2.95 \text{ A}$$ **Final answers:** - Area under curve $= 88.5$ A·ms - Mean current $= 2.95$ A