A 10-mh inductor has a parasitic series resistance of rs = 0.5 ω, as shown below. If the current is given by i(t) = 0.1 sin(20t) a, find vr(t) , vl(l), and v(t).

Respuesta :

[tex]v_r(t)=r\cdot i(t)=(0.5\,\Omega)(0.1\sin{(20t)\,A})=0.05\sin{(20t)}\,V\\\\v_l(t)=l\cdot \dfrac{d}{dt}(i(t))=(0.010\,H)(20\cdot 0.1\cos{(20t)\,A/s})=0.02\cos{(20t)\,V}[/tex]

Then the total voltage is ...

[tex]v(t)=(0.05\sin{(20t)}+0.02\cos{(20t)})\,V\approx 0.05385\sin{(20t+0.3805)\,V}[/tex]

The magnitude of the resultant voltage is √(.02²+.05²); the angle is arctan(.02/.05).