Three pieces of copper wires of lengths in the ratio 2:3:4 and with diameters in the ratio 4:5:6 are connected in parallel. Find the current in each branch if the main current is 5 A.

Let,
l₁, l₂, and l₃ be lengths of three copper wires.
D₁, D₂ and D₃ be their diameters and,
A₁, A₂, A₃ be their area of cross section. 

Given,  

l₁: l₂: l₃ = 2: 3: 4

i.e.,  l₁ = 2 l, l₂ = 3 l and l₃ = 4 l.

Also, given,

D₁: D₂: D₃ = 4: 5: 6

∴ A₁: A₂:A₃ = (4)²: (5)²: (6)² = 16: 25: 36
That is,
A₁ = 16 A, A₂ = 25 A and A₃ = 36 A

If ρ is the resistivity of copper, then $$\begin{gathered} {\mathrm{R}}_1 = \frac{{\mathrm{\rho l}}_1}{{\mathrm{A}}_1} = \frac{\mathrm{\rho } \times 2 \mathrm{l}}{16 \mathrm{A}} = \frac{1}{8}\frac{\mathrm{\rho l}}{\mathrm{A}} \\ {\mathrm{R}}_2 = \frac{{\mathrm{\rho l}}_2}{{\mathrm{A}}_2} = \frac{\mathrm{\rho } \times 3\mathrm{l}}{25 \mathrm{A}} = \frac{3}{25}\frac{\mathrm{\rho l}}{\mathrm{A}} \end{gathered}$$
and,  
${\mathrm{R}}_3 = \frac{{\mathrm{\rho l}}_3}{{\mathrm{A}}_3} = \frac{\mathrm{\rho }4\mathrm{l}}{36\mathrm{A}} = \frac{1}{9}\frac{\mathrm{\rho l}}{\mathrm{A}}$ 

$\therefore$  $$\begin{gathered} {\mathrm{R}}_1:{\mathrm{R}}_2:{\mathrm{R}}_3 = \frac{1}{8}:\frac{3}{25}:\frac{1}{9} \\ = 25 \times 9:3 \times 8 \times 9:8 \times 25 \end{gathered}$$ 

or ${\mathrm{R}}_1:{\mathrm{R}}_2:{\mathrm{R}}_3= 225 : 216 : 200$ 

$\therefore$ $$\begin{gathered} {\mathrm{R}}_1 =225 \mathrm{R}, \\ {\mathrm{R}}_2 = 216 \mathrm{R} \mathrm{and}, \\ {\mathrm{R}}_3 = 200 \mathrm{R} \end{gathered}$$ 

Let I₁, I₂ and I₃ be the currents through the wires of resistances R₁, R₂ and R₃ respectively. Then,

${\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3 = 5$                 ...(i)

and,

${\mathrm{I}}_1\times 225 \mathrm{R} = {\mathrm{I}}_2 \times 216 \mathrm{R} = {\mathrm{I}}_3 \times 200 \mathrm{R}$

$\Rightarrow {\mathrm{I}}_1\times 225 = {\mathrm{I}}_2 \times 216 = {\mathrm{I}}_3 \times 200$

$\therefore$ ${\mathrm{I}}_2 = \frac{225 {\mathrm{I}}_1}{216} = 1.04 {\mathrm{I}}_1$   
and,
    ${\mathrm{I}}_3 = \frac{225 {\mathrm{I}}_1}{200} = 1.125 {\mathrm{I}}_1$

Putting these values in equation (i) we get, 

I₁ + 1.04 I₁ + 1.125 I₁ = 5
on solving, we get
 I₁ = 1.58 A
 I₂ = 1.04 x 1.58 = 1.64 A
and, 
 I₃ = 1.125 x 1.58 = 1.78 A.