A single-core cable has a designed current of 12 A and voltage drop of \( 24 \mathrm{mV} / \mathrm{A} / \mathrm{m} \). If a maximum voltage drop of \( 3.5 \% \) is allowed, find what is the longest cable can be used if the supply is at 230 V ? 27.951 m 279.5 m 798.6 m 2795.18 m

To determine the maximum length of the cable that can be used without exceeding the allowed voltage drop, follow these steps: 1. **Given Data:** - **Current (I):** 12 A - **Voltage drop per amp per meter (\( V_d \)):** \( 24 \, \text{mV/A/m} \) - **Supply Voltage (V):** 230 V - **Maximum allowed voltage drop percentage:** 3.5% 2. **Calculate the Maximum Allowed Voltage Drop:** \[ \text{Maximum Voltage Drop} = 0.035 \times 230\, \text{V} = 8.05\, \text{V} \] 3. **Determine Voltage Drop per Meter:** \[ \text{Voltage Drop per Meter} = I \times V_d = 12\, \text{A} \times 24\, \text{mV/A/m} = 288\, \text{mV/m} = 0.288\, \text{V/m} \] 4. **Calculate the Maximum Length of the Cable:** \[ \text{Maximum Length} = \frac{\text{Maximum Voltage Drop}}{\text{Voltage Drop per Meter}} = \frac{8.05\, \text{V}}{0.288\, \text{V/m}} \approx 27.951\, \text{meters} \] **Answer:** 27.951 m