1 Makerere University Department of Electrical and Computer Engineering B.Sc. in Electrical Engineering, B.Sc. in Computer & Communications Engineering and B.Sc. Biomedical Engineering EMT^1101 – ENGINEERING MATHEMATICS I Coursework Set 1 Instructions: a) Work in groups of 2–3 members. b) Each group must include members from different programmes (BELE, BCCE, BBI). c) Clearly list each member’s name, registration number, and programme on the cover page. d) Solutions may be handwritten or typed. e) Submission deadline: 30th October 2025 at 8:00 AM (strictly). 1. For power system engineers, it is essential to ensure maximum power transfer from the source to the load. Figure 1 shows a circuit in which a non-ideal voltage source is connected to a variable load resistor with resistance 𝑅𝐿. The source voltage is 𝑉 and its internal resistance is 𝑅𝑆. Calculate the value of 𝑅𝐿 which results in the maximum power being transferred from the voltage source to the load resistor. 6 Marks Figure 1 2. Find the dimensions of the right-circular cylinder of largest volume that can be inscribed in a sphere of radius R. 4 Marks 3. Sketch graphs of the functions 2 (i) 𝑦 = π‘₯2βˆ’π‘₯βˆ’6 π‘₯+ 1 5 Marks (ii)𝑦 = π‘₯βˆ’1 π‘₯2βˆ’4 5 Marks 4. Given the system of linear equations 4π‘₯ βˆ’ 5𝑦 + 7𝑧 = βˆ’14 9π‘₯ + 2𝑦 βˆ’ 3𝑧 = 47 π‘₯ βˆ’ 𝑦 βˆ’ 5𝑧 = 11 Solve the equation using (i) Crammer’s rule 5 Marks (ii)Gauss elimination method 5 Marks 5. (a) Sketch graphs of the following functions (i) 𝑓(π‘₯) = π‘₯ |π‘₯| 2 Marks (ii) 𝑓(π‘₯) = √4 βˆ’ π‘₯2 2 Marks (iii)𝑓(π‘₯) = {π‘₯2, π‘₯ > 1 2, π‘₯ ≀ 1 2 Marks (b) An open box is to be made from an 8π‘π‘š Γ— 15π‘π‘š piece of sheet metal by cutting out squares with sides of length π‘₯ from each of the four corners and bending up the sides. Express the volume 𝑉 of the box as a function π‘₯, and state the domain and range of the function. 4 Marks 6. Evaluate the following integrals (i) ∫ sin^2 3π‘₯ cos 3π‘₯ 𝑑π‘₯ πœ‹ 2 ⁄ 0 4 Marks (ii) ∫ cos 2π‘₯ √7βˆ’3 sin 2π‘₯ 𝑑π‘₯ πœ‹ 4 ⁄ 0 4 Marks (iii)∫ π‘₯2 √4βˆ’3π‘₯ 𝑑π‘₯ 1 0 4 Marks (iv) ∫ √tan π‘₯ sec^2 π‘₯ 𝑑π‘₯ 4 Marks (v) ∫ π‘’π‘Žπ‘₯ cos 𝑏π‘₯ 𝑑π‘₯ 4 Marks
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Step 1
: To calculate the value of $R\_L$ which results in the maximum power being transferred from the voltage source to the load resistor, we need to find the value of $R\_L$ that makes the voltage across the load resistor equal to half of the source voltage.

\frac{R\_L}{R\_S + R\_L} V = \frac{1}{2} V
This is because maximum power transfer occurs when the load resistance is equal to the source resistance, and the voltage is divided equally across the two resistors. The voltage across the load resistor is given by:

Step 2
: Solving for $R\_L$ gives:

R\_L = \frac{1}{2} R\_S

Final Answer

The solutions to the given problems are presented in steps above.