A 20 ohm resistor with temperature coefficient 0.004/°C is connected to a 24 V supply. The temperature rises from 1.5°C to 10°C, determine: I) New resistance II) Circuit current before and after heating III) explain your answers. [9 marks]

Here's the solution for Q5.b: Given: Resistors R_1 = 8 \, , R_2 = 10 \, , R_3 = 15 \, connected in series. Supply voltage V = 420 \, V. Resistance of each resistor increases by 2.5% due to heating. I) Initial current Step 1: Calculate the total initial resistance. For resistors in series, the total resistance is the sum of individual resistances. R_initial, total = R_1 + R_2 + R_3 R_initial, total = 8 \, + 10 \, + 15 \, R_initial, total = 33 \, Step 2: Calculate the initial current. Using Ohm's Law, I = (V)/(R). I_initial = (V)/(R_initial, total) I_initial = (420 \, V)/(33 \, ) I_initial ≈ 12.727 \, A The initial current is 12.727 A. II) Final current Step 3: Calculate the total final resistance. Each resistor's resistance increases by 2.5%, so the total resistance also increases by 2.5%. R_final, total = R_initial, total × (1 + 0.025) R_final, total = 33 \, × (1.025) R_final, total = 33.825 \, Step 4: Calculate the final current. Using Ohm's Law, I = (V)/(R). I_final = (V)/(R_final, total) I_final = (420 \, V)/(33.825 \, ) I_final ≈ 12.417 \, A The final current is 12.417 A. III) Percentage change in current Step 5: Calculate the percentage change in current. The percentage change is given by I_final - I_initialI_initial × 100\%. Percentage change = (12.416851... \, A - 12.727272... \, A)/(12.727272... \, A) × 100\% Percentage change = (-0.310421... \, A)/(12.727272... \, A) × 100\% Percentage change ≈ -0.0243899 × 100\% Percentage change ≈ -2.439\% The negative sign indicates a decrease in current. The percentage change in current is -2.439\%. What's next?