Split phase motor added resistance and capacitance

To send the same current in both windings with a 90° phase shift, the auxiliary branch impedance must have the same magnitude as the main winding impedance but an angle that is 90° less. A resistance alone cannot shift the current by 90°, so both a resistance and a capacitor are added in series with the auxiliary winding.

Given: $f = 50$ Hz, $Z_m = Z_a = 5 + j20\ \Omega$ (before adding anything).

Main winding

$$\begin{aligned} Z_m &= 5 + j20 = 20.616\angle 75.96^\circ\ \Omega \end{aligned}$$

So $I_m$ lags $V$ by $75.96^\circ$.

Required auxiliary impedance

For $|I_a| = |I_m|$ and $I_a$ leading $I_m$ by $90^\circ$:

$$\begin{aligned} \theta_a &= 75.96^\circ - 90^\circ = -14.04^\circ \\ Z_a' &= 20.616\angle -14.04^\circ \\ &= 20.616(\cos 14.04^\circ - j\sin 14.04^\circ) \\ &= 20 - j5\ \Omega \end{aligned}$$

Let $R$ and $X_C$ be added in series: $Z_a' = (5 + R) + j(20 - X_C)$.

a) Resistance to be added

$$\begin{aligned} 5 + R &= 20 \\ R &= 15\ \Omega \end{aligned}$$

b) Capacitance to be added

$$\begin{aligned} 20 - X_C &= -5 \;\Rightarrow\; X_C = 25\ \Omega \\ C &= \frac{1}{2\pi f X_C} = \frac{1}{2\pi \times 50 \times 25} \\ &= 127.3 \times 10^{-6}\ \text{F} \end{aligned}$$

Check

$|Z_a'| = \sqrt{20^2 + 5^2} = 20.616\ \Omega = |Z_m|$, so the currents are equal; angle difference $= 75.96^\circ - (-14.04^\circ) = 90^\circ$.

        I_a (leads V by 14.04 deg)
         ^
         |   /
         |  /
         | /
  -------+-----------------> V
          \
           \ 75.96 deg
            \
             v I_m   (I_a and I_m 90 deg apart)

Answer: add $R = 15\ \Omega$ and $C \approx 127.3\ \mu\text{F}$ ($X_C = 25\ \Omega$) in series with the auxiliary winding.