contestada

A certain element has a phasor voltage of V=200∠30∘VV=200∠30∘V and current of I=2∠120∘AI=2∠120∘A. The angular frequency is 500 rad/srad/s. Determine the nature of the element.

Respuesta :

erturkmemmedli

Answer:

Pure Capacitance

[tex]C=2*10^{-5}\:F[/tex]

Explanation:

[tex]\omega=500\:\:rad/s,\:\theta_V=30^{\circ },\:\theta_I=120^{\circ }[/tex]

[tex]\theta_{VI}=|\theta_V-\theta_I|=90^{\circ}[/tex]

Notice that phasor I leads phasor V by 90 degrees.

Hence we are looking for the impedance [tex]Z_C[/tex] which is a pure capacitance.

[tex]Z_C=\frac{V_C}{I_C}= \frac{200\angle 30^\circ }{2\angle 120^\circ } =100\angle {-90^\circ }\\\\Z_C=\frac{1}{\omega C}\angle {-90^\circ }=100\angle {-90^\circ }\\\\C=\frac{1}{100\omega} =\frac{1}{100*500} =2*10^{-5}\:F[/tex]

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