# AC Circuits Current Divider Rule

The basic format for the current divider rule in ac circuits is exactly the same as that for dc circuits; that is, for two parallel branches with impedances $Z_1$ and $Z_2$ as shown in Fig. 1,
Fig. 1: Applying the current divider rule.
$$\bbox[10px,border:1px solid grey]{I_1 = {Z_2 I_T \over Z_1 + Z_2}}$$ $$\bbox[10px,border:1px solid grey]{I_2 = {Z_1 I_T \over Z_1 + Z_2}}$$
Example 1: Using the current divider rule, find the current through each impedance of Fig. 2.
Fig. 1: For Example 1.
Solution: $$\begin{split} I_R &= {Z_L I_T \over Z_R + Z_L}\\ &={(4Ω \angle 90^\circ)(20A \angle 0^\circ) \over (3Ω \angle 0^\circ) + (4Ω \angle 90^\circ)}\\ &={80A \angle 90^\circ \over 5 \angle 53.13^\circ}\\ &=16 A \angle 36.87^\circ\\ I_L &= {Z_R I_T \over Z_R + Z_L}\\ &={(3Ω \angle 0^\circ)(20A \angle 0^\circ) \over (3Ω \angle 0^\circ) + (4Ω \angle 90^\circ)}\\ &={60A \angle 0^\circ \over 5 \angle 53.13^\circ}\\ &=12 A \angle -53.13^\circ \end{split}$$