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- The circuit is connected to a 400-Hz line with an applied voltage of 35.678 V. The resistor has a true power of 14.4 W. and there are 12.96 inductive VARs and 28.8 capacitive VARs. ET35.678VITZVAPFERIRRP14.4WELILXLVARsL12.96LECICXCVARsC28.8CFind which one leads and by how much (in degrees). • v1(t) = 22 cos(377t - 30°) • i1(t) = 10 cos(377t + 20°) i1 [ leads or lags ] v1 by x degrees i1 [ leads or lags ] v1 by x degrees. ..I = - 10 - j 12. With ω = 1500 rad/s, find i(4 ms). Select one: a. – 9.68 A b. 15 angle (-130o) A c. 6.25 A d. – 12.26 A e. – 10.82 A
- delta to wye. find Vt Rt It and power. pls write your answe pls. dont type it because i don't understand when u type.A circuit consists of Xa = j5ohms, Xc= j5 ohms and R= 5 ohms all are connected in parallel. Find the equivalent impedance.USE SOURCE TRANSFORMATION METHODredraw it to understand step by step soln. answer it in 1hr.
- In the circuit, using the KIRCHOFF'S METHOD determine the current in each branch Expected answers: I1 = 0.1998<-34.58 degrees, A I2 = 0.4305 <624.09 degrees, A I3= 0.5610 <-173.62 degrees, A or 0.5610 <6.38degrees, AA circuit consists of Xa=j5 ohms , Xc=-j5 ohms and R= 5 ohms all are connected in parallel. Find the equivalent impedance.From step1, how did you get that 50+(80*j60/80+j60) = 78.8+j38.4? also in step 2, how did you get that 5/(0.0125-0.0167) = 143.63 + j191.89?
- The voltage and current at the terminals of the circuit element are zero for t<0. For t≥0 they are v=50e−1600t−50e−400t V,i=5e−1600t−5e−400t mA. Find the power at t=625 μs.V:16) Find Iy , Rx = 5Ωans: a. i = 10 cis-30° Ab. v = 220 cis25° Vc. I = 7.0711 cis-30° Ad. V = 155.5635cis25° Ve. Z = 22 cis55° Ω