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in Physics by (64.8k points)

A cable is 300 km long and has a conductor of 0.5 cm in diameter with an insulation covering of 0.4 cm thickness. Calculate the capacitance of the cable if relative permittivity of insulation is 4.5.

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Capacitance of a cable is C = (0.024εr)/log10(b/a)μF/km

Here, a = 0.5/2 = 0.25 cm ; b = 0.25 + 0.4 = 0.65 cm ; b/a = 0.65/0.25 = 2.6 ; log10 2.6 = 0.415

∴  C = (0.024 x 4.5)/0.415 = 0.26

Total capacitance for 300 km is = 300 × 0.26 = 78μF

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