Question

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.

Answer 1

Pure Capacitance

`C=2*10^(-5)\:F`

Step-by-step explanation:

`\omega=500\:\:rad/s,\:\theta_V=30^(\circ ),\:\theta_I=120^(\circ )`

`\theta_(VI)=|\theta_V-\theta_I|=90^(\circ)`

Notice that phasor I leads phasor V by 90 degrees.

Hence we are looking for the impedance
`Z_C` which is a pure capacitance.

`Z_C=(V_C)/(I_C)= (200\angle 30^\circ )/(2\angle 120^\circ ) =100\angle {-90^\circ }\\\\Z_C=(1)/(\omega C)\angle {-90^\circ }=100\angle {-90^\circ }\\\\C=(1)/(100\omega) =(1)/(100*500) =2*10^(-5)\:F`