Learn everything about Current Division in Parallel Circuits.

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:
current-divider-design-problem
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  Rin 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: R= 1.5 R1

From formula of equivalent resistance.
1/Req =  1/R1 + 1/R2.

1/100 =  1/R1 + 1/R2.

By putting R= 1.5 R1 we achieve R= 166 Ω
and R= 250 Ω