Answer to Question #287575 in Electrical Engineering for Saiteja

Question

A current of 90 A is shared by three resistances in parallel. The wires are of the

same material and have their lengths in the ratio 2: 3: 4 and their cross-sectional areas

in the ratio 1: 2: 3. Determine current in each resistance

Answer 1

The voltage across all resistances is the same:

$V=I_1R_1=I_2R_2=I_3R_3.$

By Kirchhoff's current law:

$I_1+I_2+I_3=90\text{ A}.\\\space\\ \dfrac V{R_1}+\dfrac V{R_2}+\dfrac V{R_3}=90\text{ A}.$

The resistance of each resistor is

$R_1=\dfrac{\rho L_1}{\pi r_1^2},\\\space\\ R_2=\dfrac{\rho L_2}{\pi r_2^2}=\dfrac{\rho (\frac 32L_1)}{\pi (2r_1)^2}=\dfrac 38 R_1,\\\space\\ R_3=\dfrac{\rho L_3}{\pi r_3^2}=\dfrac{\rho (2L_1)}{\pi (3r_1)^2}=\dfrac 29 R_1.$

Since $I_1R_1=I_2R_2=I_3R_3,$

$I_1=\dfrac 38I_2=\dfrac29I_3,\\\space\\ I_1+\dfrac 83 I_1+\dfrac 92I_1=90,\\\space\\ I_1=11\text{ A},\\ I_2=29\text{ A},\\ I_3=50\text{ A}.$