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Capacitor for quadrature currents (4.5+j3.5 Ω)

Short answer

Main winding angle = tan⁻¹(3.5/4.5) = 37.87°. For quadrature, starting-branch angle must be −52.13°, yielding Xc = 15.71 Ω. Capacitance C = 1/(2π×50×15.71) ≈ 202.6 μF.

A 250 W, 230 V, 50 Hz, single-phase capacitor start induction motor has the following constants for its main and starting windings: Zm = (4.5+j3.5) Ω and Zs = (9.5+j3.5) Ω. Determine the value of the starting capacitor that will place the main and starting winding currents in quadrature at starting.

For the currents to be in quadrature, the auxiliary (starting) winding current must lead the main winding current by 90∘90^\circ. So the angle of the starting branch impedance must be 90∘90^\circ less than that of the main winding.

Given: Zm=4.5+j3.5 ΩZ_m = 4.5 + j3.5\ \Omega, Zs=9.5+j3.5 ΩZ_s = 9.5 + j3.5\ \Omega, V=230V = 230 V, f=50f = 50 Hz.

Main winding angle

θm=tan⁡−13.54.5=37.87∘(Im lags V by 37.87∘)\theta_m = \tan^{-1}\frac{3.5}{4.5} = 37.87^\circ \quad (I_m \text{ lags } V \text{ by } 37.87^\circ)

Required angle of starting branch

θs=37.87∘−90∘=−52.13∘\theta_s = 37.87^\circ - 90^\circ = -52.13^\circ

With capacitor: Zs′=9.5+j(3.5−XC)Z_s' = 9.5 + j(3.5 - X_C), so

3.5−XC9.5=tan⁡(−52.13∘)=−4.53.5=−1.28573.5−XC=−12.214XC=15.714 Ω\begin{aligned} \frac{3.5 - X_C}{9.5} &= \tan(-52.13^\circ) = -\frac{4.5}{3.5} = -1.2857 \\ 3.5 - X_C &= -12.214 \\ X_C &= 15.714\ \Omega \end{aligned}

Capacitance

C=12πfXC=12π×50×15.714=202.6×10−6 F\begin{aligned} C &= \frac{1}{2\pi f X_C} = \frac{1}{2\pi \times 50 \times 15.714} \\ &= 202.6 \times 10^{-6}\ \text{F} \end{aligned} 
   o----+-------------+
        |             |
     Z_m=4.5+j3.5   Z_s=9.5+j3.5
        |             |
 230 V  |            [C]  ~203 uF
        |             |
        |            CS (centrifugal switch)
   o----+-------------+

Check: Zs′=9.5−j12.214=15.47∠−52.13∘Z_s' = 9.5 - j12.214 = 15.47\angle -52.13^\circ; angle between currents =37.87∘+52.13∘=90∘= 37.87^\circ + 52.13^\circ = 90^\circ.

Answer: starting capacitor C≈202.6 μFC \approx 202.6\ \mu\text{F} (XC=15.71 ΩX_C = 15.71\ \Omega).

✓ Answer checked by Claude Sonnet 5.5

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