How do electrical cables work?

Electrical cables are crucial components in electrical systems, designed to transmit electrical power or signals from one point to another. Their operation relies on the principles of electrical conductivity and insulation.

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Key Components of an Electrical Cable

• Conductor: The core of the cable, typically made of high-conductivity materials like copper or aluminum. It allows the flow of electric current due to free electrons.
• Insulation: A non-conductive material (e.g., PVC, rubber) that surrounds the conductor, preventing short circuits and ensuring safety by containing the current.
• Sheathing: Provides mechanical protection and shields the cable from environmental factors like moisture, heat, or physical damage.
• Shielding (Optional): Used in communication cables to reduce electromagnetic interference (EMI), often made of braided copper or aluminum foil.

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How Electrical Cables Work

• Electric Field Generation: A voltage is applied across the cable's conductor, creating an electric field that drives the flow of electrons.
• Current Flow: Free electrons in the conductor move under the influence of the electric field, resulting in electric current.
• Energy Transfer: The cable transfers energy from the power source to the load (e.g., a motor, light bulb, or electronic device).
• Heat Dissipation: As current flows, resistance in the conductor causes heat generation. Insulation prevents heat loss and minimizes the risk of overheating.

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Example: Power Transmission in a Building

Given:

• Cable length (L): 30 m,
• Conductor material: Copper (ρ = 1.68 × 10⁻⁸ Ω·m),
• Cross-sectional area (A): 1.5 mm² (A = 1.5 × 10⁻⁶ m²),
• Voltage supplied (V): 230 V,
• Power of the bulb (P): 100 W.

Step 1: Calculate the Resistance of the Cable:
R = ρ * (L / A)R = (1.68 × 10⁻⁸) * (30 / 1.5 × 10⁻⁶) = 0.336 Ω
Step 2: Determine the Current Flow:
I = P / VI = 100 / 230 ≈ 0.435 A
Step 3: Calculate the Voltage Drop:
V_d = I * R_totalV_d = 0.435 * 0.672 = 0.292 V
Step 4: Calculate the Heat Loss:
P_loss = I² * R_totalP_loss = (0.435)² * 0.672 ≈ 0.127 W
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Conclusion

The cable efficiently delivers power to the bulb with minimal voltage drop (0.13%) and low power loss (0.127 W). This example highlights how cables work to transfer energy while accounting for resistance and efficiency.