An AC circuit has a resistor, capacitor and inductor in series with a 120 V, 60 Hz voltage source. The resistance of the resistor is 3 Ohms, the inductance of the inductor is 0.01 Henry; the capacitance of the capacitor is 0.00075 Farads. Find the following: (i)-Reactance of the capacitor and inductor (ii)-Total impedance Current (iii)-Apparent power Real power vi) Reactive power vii) Power factor angle viii) Power factor.

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aristeus aristeus · Answer 1 · 2019-09-03T13:02:42+02:00

Answer:

(i) 3.5385 ohm, 3.768 ohm (ii) 39.89 A (III) 4773.857 W (vi) 348 var (vii) 0.9973 (viii) 4.1796°

Explanation:

We have given voltage V =120 volt

Frequency f=60 Hz

Resistance R =3 ohm

Inductance L =0.01 H

Capacitance C =0.00075 farad

(i) reactance of of inductor [tex]X_L=\omega L=2\pi fL=2\times 3.14\times 60\times 0.01=3.768ohm[/tex]

[tex]X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 60\times 0.00075}=3.5385ohm[/tex]

(ii) Total impedance [tex]Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{3^2+(3.768-3.5385)^2}=3.008ohm[/tex]

Current [tex]i=\frac{V}{Z}=\frac{120}{3.008}=39.89A[/tex]

(viii) power factor [tex]cos\Phi =\frac{R}{Z}=\frac{3}{3.008}=0.9973[/tex]

(VII) [tex]cos\Phi =0.9973[/tex]

[tex]\Phi =4.1796^{\circ}[/tex]

So power factor angle is 4.1796°

(iii) Apparent power [tex]P=VICOS\Phi =120\times 39.89\times 0.9973=4773.875W[/tex]

(vi) Reactive power [tex]Q=VISIN\Phi =120\times 39.89\times SIN4.17^{\circ}=348var[/tex]