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.

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 = {R

_{1}/ (R_{1}+ R_{2})} * 2 A
0.005 R

_{1}+ 0.005 R_{2}= 2 R_{1}
or alternatively

R

_{2}= 399 R_{1}... (1)
From Ohm's law:

R

_{2}= V/I_{2}= 5 V / 5 mA = 1 kΩ
Putting the value of R

_{2 }in equation 1
R

_{1}= 2.50 Ω
So a current divider whose design involves R

_{1}= 2.50 Ω and R_{2}= 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

_{2 }= 1.5 R_{1}
From formula of equivalent resistance.

1/R

_{eq }= 1/R_{1}+ 1/R_{2}.
1/100 = 1/R

_{1}+ 1/R_{2}.
By putting R

_{2 }= 1.5 R_{1}we achieve R_{1 }= 166 Ω
and R

_{2 }= 250 Ω