Question

Julius D. Cagampang's question:

This is a solving problem with the use of the differential equations. Help me to answer this. Thank you.

Circuits In the RL-series circuit shown in Figure 1.14, assume that V(t) = V, sin wt and I(0) = 0. Show that Vowl I(t) Ve e-RT/L+ sin (wt - o), R? + w?L? (R? + w?L?)12 = = where 0 is an angle defined by wL R sin cos (R? + w?L?)12) (R+ w?L?)12 Note that the current 1(1) is the sum of two terms; as 1 →00, the first term dies out and the second term dominates. For this reason the first term is called the transient and the second term the steady state. R + V(0) I(1) Figure 1.14

Answer 1

+ Apply KVE) + + VE) ti I(t) - V(6) + ICE) R + VL = 0 – VCE) +I(t) B+ L d[(t)=0 dt dict) TE + B ICC) VCE) Vo Sinwt complementary Solution dict) de = + ĢI(t) (0+2) I(t) =0 - Da Ict)= 4c (t) = ce è het ..' c = constant. x Particular solution, ( DRL) Vo sinut 4p(t) = O-RIL) (O+B) Vo CO-RIL) Siswt D-RIL) replace or will con Vo CD-RIL) sincot 4plt) = -w"_(RIL)" Coswt.W- & siswt yo [cos t 1) - wn-(R/L)

Up (t) = [vow Coswt - Voß sinwt cort A/ VO WL Coswt - Vo R sinwt ] conftar , JPCE) - VORSinot VO WL Coswt (WN+R")l2 Carthylla 1 VOR Colth WYPRY!l2 Sinwt Vocot cosot (GMT+B")'12 1 → sinut +R)/? (ww + R) 1/2 → VO COBQ Sinat - (COMT +R")'l2 Sind Coswt Vo → sin (wt-d) (w+BU12 Vo COF COSwt VOR Siscot 4p(t) = Ca'nº+B) general solution »ICH4Ct) = c Ct) + 4pct) -ht ICE)= Ce + VOR Sinct -VOWL Coswt w*+B)

ICO) = 0 = -O(RIM) ce + VOR sinw (0) - VOWL Cosw10) RAY O = C NowL Vowl (WLYR") C = +R) e Rt/L Vowt I(t) = VOR Siswt-VOCf cosct Corentar CwYn+RV) From the ego ©, we can We can represent as Vook But vo ICE) = e CCM + RV Sin Cwt-0) (comentar)/2