# Answer to Question #142183 in Electrical Engineering for Thuso

Question #142183
3.1 A copper and an aluminium conductor are connected in parallel to supply a load
of 300A over a distance of 500m. The two conductors have equal cross-sectional
areas of 50mm2. Take the resistivity of coper as 0.018µΩm and that of aluminium as
0.028µΩm and calculate
3.1.1 the total resistance of the two conductors (9)
3.1.2 The voltage drop across the compound line (3)
3.1.3 The current carried by each conductor (6)
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2020-11-03T15:40:06-0500

Solution: Following are the datas given,

Length of Aluminium conductor( "l_a" ) = Length of copper conductor ( "l_c" ) = 500m

Area of cross section of Aluminium conductor( "A_a" ) = Area of cross section of copper conductor ( "A_c" ) = 50mm2. = "50\\times{10}^{-6}" m

Resistivity of copper ( "\\rho_c" ) = 0.018"\\mu\\varOmega m"

Resistivity of aluminium( "\\rho_a" ) = 0.028"\\mu\\varOmega m"

The total load current = 300A.

(3.1.1)There fore,

Total resistance of Copper conductor = "\\rho_c \\times l_c\/ A_c" = "0.018\\times10^{-6} \\times500\/(50\\times10^{-6})"

"R_c" = "0.18\\varOmega"

Total resistance of Aluminium conductor = "\\rho_a \\times l_a\/ A_a" = "0.028\\times10^{-6} \\times500\/(50\\times10^{-6})"

"R_a" = "0.28\\varOmega"

Total resistance of two conductors will be "(R_t)" = "R_a \\times R_c\/(R_a + R_c)" ="0.1096\\varOmega"

(3.1.2) Voltage drop across the compound line = "R_t\\times" Load current

= "0.1096\\times300 = 32.87V"

(3.1.3) Current carried by aluminium conductor = "R_c \\times Load current\/(R_a + R_c)"

"0.18\\times300\/0.46 = 117.39 A"

Current carried by Copper conductor = "R_a \\times Load current\/(R_a + R_c)"

"0.28\\times300\/0.46 = 182.61 A"

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