The direct current source is located at a distance of 10 meters from an electric heater with an electrical resistance of the coil of 10 ohms. What is the minimum allowable cross-sectional area of the aluminum wire connecting the heater to the current source, provided that the voltage drop across the supply wires should not exceed 2% of the voltage across the heater coil?
Let current I flow through the circuit. Voltage drop across the spiral U = I * Rс, where Rс is the resistance of the spiral.
The voltage drop across wires with resistance Rп should be 2% of U:
0.02 * U = I * r
Instead of U, we substitute I * Rс:
0.02 * I * Rс = I * Rп.
We find that Rp = 0.02Rs.
Rп = (r * L) / S (r is the resistivity of aluminum, L is the length of the wires, S is the cross-sectional area).
The length of the wires is equal to twice the distance – L = 20 m.
S = (r * L) / Rp = (r * L) / 0.02Rc = (0.028 Ohm * mm² / m * 20 m) / (0.02 * 10 Ohm) = 2.8 mm².
Answer: 2.8 mm².
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