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]

For resistors in series, the total resistance is the sum of individual resistances.

$$R_{initial, total} = R_1 + R_2 + R_3$$ $$R_{initial, total} = 8 \, \Omega + 10 \, \Omega + 15 \, \Omega$$ $$R_{initial, total} = 33 \, \Omega$$

Using Ohm's Law, $I = \frac{V}{R}$.

$$I_{initial} = \frac{V}{R_{initial, total}}$$ $$I_{initial} = \frac{420 \, V}{33 \, \Omega}$$ $$I_{initial} \approx 12.727 \, A$$

The initial current is $\boxed{\text{12.727 A}}$.

II) Final current

Each resistor's resistance increases by 2.5%, so the total resistance also increases by 2.5%.

$$R_{final, total} = R_{initial, total} \times (1 + 0.025)$$ $$R_{final, total} = 33 \, \Omega \times (1.025)$$ $$R_{final, total} = 33.825 \, \Omega$$

Using Ohm's Law, $I = \frac{V}{R}$.

$$I_{final} = \frac{V}{R_{final, total}}$$ $$I_{final} = \frac{420 \, V}{33.825 \, \Omega}$$ $$I_{final} \approx 12.417 \, A$$

The final current is $\boxed{\text{12.417 A}}$.

III) Percentage change in current

The percentage change is given by $\frac{I_{final} - I_{initial}}{I_{initial}} \times 100\%$.

$$\text{Percentage change} = \frac{12.416851... \, A - 12.727272... \, A}{12.727272... \, A} \times 100\%$$ $$\text{Percentage change} = \frac{-0.310421... \, A}{12.727272... \, A} \times 100\%$$ $$\text{Percentage change} \approx -0.0243899 \times 100\%$$ $$\text{Percentage change} \approx -2.439\%$$

The negative sign indicates a decrease in current.
The percentage change in current is $\boxed{\text{-2.439\%}}$.