Current Divider Design Problem

Today you'll learn to design a current divider circuit using basic current divider rule.

Design a current divider which attenuates 2 A current to 5 mA. The operating voltage is 5 V.

Let's draw the circuit of current divider:

For our present case the formula will become:

5 mA = {R₁ / (R₁ + R₂)} * 2 A

0.005 R₁ + 0.005 R₂ = 2 R₁

or alternatively

R₂ = 399 R₁... (1)

From Ohm's law:

R₂ = V/I₂ = 5 V / 5 mA = 1 kΩ

Putting the value of  R₂ in equation 1

 R₁ = 2.50 Ω

So a current divider whose design involves R₁ = 2.50 Ω and R₂ = 1 kΩ will dissipate 2 A to 5 mA.

Current Divider Design Problem 2

Design a divider which divides input current to the ratio of 3:2. So that that the equivalent resistance is 100 Ω.

The input is 3 + 2 = 5, lets assume it in amperes.

Now apply the formula with 5 A input and 2 A as output for R2 we obtain: R₂ = 1.5 R₁

From formula of equivalent resistance.

1/R_eq =  1/R₁ + 1/R₂.

1/100 =  1/R₁ + 1/R₂.

By putting R₂ = 1.5 R₁ we achieve R₁ = 166 Ω

and R₂ = 250 Ω