Question 2

The Correct Answer is C ### Explanation of Why Option C (18.1 amps) is Correct The maximum current allowed to be drawn from the secondary winding of a transformer is determined by the transformer's power rating (Apparent Power, $S$) and the voltage across the secondary winding ($V_s$). 1. **Determine the Secondary Voltage ($V_s$):** * The transformer is a step-down transformer. * Primary Voltage ($V_p$): 440 V * Turns Ratio ($N_p : N_s$): 4 : 1 * The secondary voltage is calculated by: $$V_s = V_p \times \frac{N_s}{N_p}$$ $$V_s = 440 \text{ V} \times \frac{1}{4}$$ $$V_s = 110 \text{ V}$$ 2. **Calculate the Maximum Secondary Current ($I_s$):** * The power rating ($S$) is 2 kVA (2000 VA). * The relationship between apparent power, voltage, and current is $S = V_s \times I_s$. * The maximum current is: $$I_s = \frac{S}{V_s}$$ $$I_s = \frac{2000 \text{ VA}}{110 \text{ V}}$$ $$I_s \approx 18.18 \text{ amps}$$ Rounding to one decimal place, the maximum current allowed to be drawn is **18.1 amps**. --- ### Why Other Options Are Incorrect **A) 1.1 amps:** This value is approximately the primary current ($I_p$). $$I_p = \frac{2000 \text{ VA}}{440 \text{ V}} \approx 4.54 \text{ amps}$$ If 1.1 amps were the primary current, the power rating would only be $440 \times 1.1 = 484 \text{ VA}$. This option is far too low for a 2 kVA transformer. **B) 4.5 amps:** This value is the maximum current allowed on the primary side ($I_p$). $$I_p = \frac{2000 \text{ VA}}{440 \text{ V}} \approx 4.54 \text{ amps}$$ Since this is a step-down transformer (voltage decreases), the current must increase in the secondary, making the primary current the wrong answer. **D) 22.7 amps:** This value results from incorrectly using the primary voltage (440 V) with a different power rating, or incorrectly calculating the secondary voltage. For instance, if one assumed $V_s$ was 88 V (a ratio of 5:1), then $2000/88 \approx 22.7$ amps. However, based on the $4:1$ ratio, the correct secondary voltage is 110 V, making this option too high.