9.1 The electric welding machine is specified as 5,3 kW, 220 V - NSC Technical Sciences - Question 9 - 2023 - Paper 1

To calculate the resistance (R) of the electric welding machine, we use the power formula:

$P = I^2R$

Given:

• Input Power, P = 5.3 kW = 5.3 \times 10^3 W
• Input Current, I = 24.3 A

Rearranging the formula for R gives: $R = \frac{P}{I^2}$

Substituting the values: $R = \frac{5.3 \times 10^3}{(24.3)^2} \approx 8.98 \; \Omega$

To find the energy dissipated (W), we use:

$W = P \times t$

Where:

• P = 5.3 kW = 5.3 \times 10^3 W
• t = 30 minutes = 30 \times 60 = 1800 seconds

Calculating the energy: $W = 5.3 \times 10^3 \times 1800 \approx 9.54 \times 10^6 \; J$

First, convert the energy calculated in kWh:

$E = \frac{W}{3600} = \frac{9.54 \times 10^6}{3600} \approx 2.65 \; kWh$

Now, calculate the cost: $\text{Cost} = E \times \text{unit price} = 2.65 \times 0.75 \approx R1.99$