Question

Vs(t) (+) 0.1H Zeq If Vs(t) = 10 cos (10t - 90°) V, determine:

If Vs(t) = 10 cos (10t - 90o) V,

determine: brief description and legible formulas and variables
Step 1: The voltage phasor VSF
Step 2: The equivalent impedance Zeq
Step 3: The phasor current IF
Step 4: The steady state current I(t)ss
Step 5: The complex power (S = ½VSFIF* = P + jQ).
Step 6: The average (P) and reactive power (Q).
Step 7: The power factor and sketch of the power triangle.

Answer 1

solni- Nst) CO I) 30.1 H Given Vs(t) = 10 cos(106-90°) V | = 10 Cor Go'-lot) v | Sinle Cost) Caso step 1: The voltage phasor VSE Vs (t) - 10 Cos go'-lot)v . = 10 sin10t8 =10 AVst) = 10 sin10t ^ } = 10 난나 아 95 ㅜ wt

Step 2: The equivalent impedance zeel Zeq> W R2+x? _tante R- de , L= 0,11 XL: 271f1= 2,74% 10, X0.1 25 = 12 Reqlz Ni+12 = ar 0- tan (d) - tant (4): 45 Z= n2 Last Phasor diagram zeg/ x, XLz1s . 02450 0 Rela

step3: Phasor current IF IF: V 10 sin 10+ Z 1tj = 10 sin 100 2 L45 = 10 sin (106-450) = 552 sin (10t - 450) 5.7.04 sin (107-45) risin Clot-450) uit ' V q r 10 sin lot Vengr 4.09 sin llot-450) Wć 02450 step 4: steady state cument Iss In Laplace domein Lo (vset)) z ál Roit lde)

L (10 sin lot), Iv [R+S4] 10X10 satice z IS) [1+0.13] I(S) z 100 (1+0.15) 84+100) z 1000 (5+10) (5271000) LITAR lim I(t) #to lim s Is) - sto Sto 2. z lem 1000s; . . sto (5+10) (S2+1000) z 0 - steady stale curent complex power E osv. I* step 5 TESTIR

= { x 10 siniot.(4-04 sin(106-45-))* = {x10x4.07.ejo", e145 = 25v2 ej 45 = 2512 ( cos 45 +jsin 450) = 2502 ( 4 /3 + 2) z 25 (1+j) 2. 25+25jz Complex step 6: Average power P = Re (25+253), 25W Reactive power P = Im (25+25;) z 256 W power

stepI: Power factors Cos 45= 0,709 lagging Power triangle IX, zin 10 - R=12