# Problems solving in electrical engineering – Current divider

General scheme of **the Current divider** is shown in Fig.1.

Fig.1 –General scheme of **the current divider**

The scheme shows that the current I is divided on resistances into currents I1, I2, I3, …, In (n is the number of resistors).

We know the current I and resistors resistances. It is necessary to find currents I1, I2, I3, …, In.

We make general formula, allowing to find any of these currents.

To find the current I we replace all resistors with equivalent resistors (Fig. 2).

Fig.2 – Equivalent circuit of resistors

We find the resistance of the equivalent resistor from the formula of parallel connection of resistances:

(1)

Voltage U is applied to the resistance of the equivalent resistor, hence, through this resistor the current I will flow, which can be found by Ohm’s law:

(2)

We know the values of the current I and the resistance of the equivalent resistor, so we express U from the formula (2) and substitute the equivalent resistor from the formula (1):

(3)

Make general formula with the help of which any current in any branch can be found.

We find the value of the current in the branch by Ohm’s law

(4)

We know the resistance of the resistor Ri (resistor in the branch), substitute in the formula U from the formula (3):

5)

That is, according to the formula

(6)

you can find current in any branch.

In the case with 2 resistances (n=2) the formula is greatly simplified.

Scheme of the **current divider ** with two resistors is shown in Fig. 3.

Fig.3 –Scheme of **the current divider** with two resistors

Formula (6) becomes:

(6)

By this formula we find currents I1, I2:

(7)

Final formula:

(8)