Question

10 points Save Ans A sinusoidal voltage (t) has arms value of 220 V. The voltage completes one complete cycle in 20 msec (milliseconds). The magnitude of the voltage at zero time is 200 V. Find the expression for v(t) using the cosine function, v(t) = 311.12 cos(314.167 +50) ov(t) = 220 cos(314.16+ + 50") v(t) - 311.12 cos(314.167 + 24.62") v(t) - 311.12 cos(507 +50°) v(t) = 220 cos(50t + 24.62") v(t) - 311.12 cos(501 + 24.62") Moving to the next question prevents chang Question 8 of 16 e م اكتب هنا للبحت

Answer 1

Vorms Imp- grel from q us- Veren= 220 Volt, T=Romsec=20x103 sec, at oto 4 200v4 Common wave equ for cosine function - | V = Vimax Cos Cotto) | where w=ant inerasion ? opposition in Qui- . 360f in t (ast)=311-12 Cos (314.164150) (6) 10+1=220.6082314.164+503 X Opposition elemination technica Vmax-te Verms or (C) Vit 1=311.12 Cod (31416 2 + 24.62) = 1-414 x 220=311-12 Volt (d) Vit = 311-12 Cod (sott50) So, opption (b) andle False f = 1/2 = 26x103 - 1000 = 50 Hz (e) 11 = 220 cos(502 +24-62) X [W= 2XF = 314-18 radion (f) VH1 = 311-12 Cod (50t 129.62) So, opposition d) and (f) ako False Conclusion Concusion opposition fromeqro V-311-12 Cod (314.16 2 60) (a) is current Find out from que from ato Vitt / 200 Put eqr 200 = 311.12 Cos Coto) |VH) = 311-12 COM(314164 +50°) •

L pression for h asa rms value of 220 V The vo using the casine hur (t) = 311.12 cos(314.161 + 504) - 220 cos(314.161 • som VO) 311.12 cos(314.167 + 24.67"