+ VL ell Vị Problem 2: Appearing to the right is again our famil- iar RLC series circuit, this time with a new voltage la- bel, v1. Also once again, for your solving convenience, is a solution for this circuit when the voltage source is + constant. Va C i1-part (t) = 0 R SKie4it + K2eª2t, (Kjeait + K2teªit, if a1 = a2. if a1 7 a2; i1-cor 1-comp(t) i1(t) = i1-comp(t)+i1-part (t) where R + 2L' V(2L R R a2 = – 1 R 1 a1 = – and LC 2L V(2L LC Find an expression for v1(t) assuming a1 # a2. There is no need to solve for K1 and K2 in this problem. Hint: Those of you who are still unsure of the difference between a voltage drop and a voltage with respect to ground: learn it now! In terms of R, L, C, va, K1, K2, a1, and a2 # a1 the voltage v1(t) + I

In the Circuit show below, Ri= 2e Rz= 32 Use Ieplace treno Form to Find the crrent io 4) F Is (+) = 3ét u(t) CA] 22 Is (+)

A series RLC circuit containing a resistance of 1202, an inductance of 0.15H and a capacitor of 100uF are connected in series across a 100V, 50Hz supply. Calculate the total circuit impedance, the circuits current and draw the voltage phasor diagram. R = 120 www VR I L = 0.15H V₁ Vs = 100V, 50Hz C = 100uF HE Vc

Problem #4: FOR the FOLLOWING SET OF EQVATIONS CONSTRUCT this Connosponding Cincuit. V₁-V₂ V₂_Vi 10 + V3-V₁ 10-20 + V₁-V3 10-j 20 V₂ J10 + + V3-√₂ jto V₂ _Vs j 40 -20.