Question

Calculate each of the circuit parameters for the provided circuit. RT(N) = Is(mA) = V1(V) = 1(MA) = PR1MW) = V2(V) = 12(mA) = V3(V) = 13(mA) - PR2(MW) = PR3(MW) - PR4(MW) = 14(MA) - V4(V) - V5(V) = 15(mA) = PR5(W) - PE(MW) = + 1 + V3 - 13 W R (Sko) R (ko) E-40V Rz(2010) R (6) 4 W RO

Answer 1

soln)

+01 + V: - AVA m R (30) R (50) E 40V R2(100) V R.(66) 34 놀 Vc R (1)

Let vezou a Node A has potential UA Applying KLL at Node' Ñ' I, = I2 + I Iz [ke KCL - Sum af Incoming current equal to current outgoing ent I3 = Iy = Is [: R1, R24 Rz are in series] I, = EUA [ Ohm's Law w Ri I2 = VA -Vc Ohm's Law R₂ . Iz = VA-Vc ohm's Law Rzt Ry TRS Putting the value of I, I AI equation ③ we get a on VA-ve + E-VA R, VA-Vc RzfR y TRS R₂ => E-VA VA u + [:Veo] R R₂ Rz tR y TRS =) 40-VA VA 5 10 10 => VAT VATZUA - ulo 10 UVA = 80 20v VAF Via E-VA = 40-20 = 200. I,= 20 - uma. 브 Ri SK v2 = VA-Vc = 200. 20 Iz = = = 2 mA : U2 R₂ lok Iz = 20 VALVE Rz TQ y tRg - 2 MA TOK V3 - Iz R3 = 2(3) 6V = Iy = Is = I2 = 2mR. Vy = (IU) Ry = (26) = 12V Ug = (Is) RS = (2) (1) = 2u. 4MA • Is = I, = [R, is is in series with Rq = yo E = - 10K : Ig ។ • PE = Is E = (40) (4) = 16omw. • P, = I (20)(4) = 80 mwi · Pz = V2 I2 = (20)(2) = 40mw Pz = Vz I3 = (6) (2) = 12 mw • Py = Vy Iy = (12) (2) = 24 mw Pr = Us Is = 12) (2) = umw.