Question

12) Find vo in the following circuit using nodal analysis. 1 H 322 WW 10 cos(1 – 45°) V vo 5 sin(a + 30°) V (Answer: 15.73 cos (t + 247.9°) V)

How to solve this? Please show how to change the capacitor and inductance into phasor. Thanks

Answer 1

Aug Vo ДАА 1H 38 locos + Vo IF (4-45 is sin(t+30) 22 7 Generally V- Vm sin ( wtto) or v = Van lo Coefficient of t is wie angular frequency In the given question w = 1 rad/ sec. It -45) =10 Süs (90 + (7-4) 1 = 10 sui Ct+45) =10 L45 V =10 cos or v Similarly, V = 5 sun (t +30) =5 /309 1 90-/ ا= to change R, L, C into Phasoro R = Ř Lo R = 3 Loo X = X 190 => XL X = į WL =j1 = 1290 Xc =Xc L-go Хc, aІ. fac ji Now apply KCL or Nodal an alysis at node Vo (marked in above fig) vo-10L45' + vo + Vo - 5/30 IL-90 Ilgo 3 20

No 3 Lo 10 L45° + V6 290 340 + Vo L-90-5/30-90= or No 10 245 + No 1907 Vo Logo=5 260 3 3 = 0 Here, Vo Lgot Vologo of Both phasor & 1 have same magnitud but in opposite direction so numerically their sun is equal to zero. . so above equation can be written as vo 3 -5 2-60 + 10 (45° 3 No 3 = 5.243 222.108 Vo C 15.73 L-22.1 07 Vo 1 vo = 15.73 sin (wt - 22:1) 15.73 świn (360+ wt - 22:1) As [sen (360 +O) = suis ] NA - 15.73 sus (wt +337.9) o No - 15.73 cos (wt + 247.9)