2260 N. Cotter PRACTICE FINAL EXAM SOLUTION: Prob 3 3. (50 point) u(t) V i(t) L - R v(t) C - The initial energy tored in the circuit i zero. 500 Ω L 200 mh a. Chooe value of R and C to accomplih the following: () v(t) and i(t) are decaying inuoid 90 out of phae with each other. (2) /α T, where α i the exponential decay contant and T i the period of ocillation of the decaying inuoid. b. With the component value you choe in the circuit, write numerical expreion for v(t) and i(t). an: a) R 500Ω, C 32.4 µf b) v(t) 59 e -6.7t in(388t) mv i(t) 2 e -6.7t co(388t) ma ol'n: (a) Thi problem could actually be olved by olving the R problem with the general olution to the differential equation for a parallel R (obtained by taking the Thevenin equivalent of the voltage ource and two reitor). The Laplace tranform approach i a bit more traightforward, however, and would alo work for input ource other than a tep function. We begin by Laplace tranforming the ource and circuit component. Since no initial energy i tored in the circuit, we have no extra current or voltage ource in the component model. In general, we would have to include uch ource to account for the initial condition. L{u(t)} -domain model: R L R L L L C L C
V g () V I() - R V() C - L Firt, we find V(). We can then find I() from V() uing Ohm' law. Our calculation are cleaner if we ue a Thevenin equivalent for V g (), R, and. V g () - R V g () R R R2 - R R2 Note: After we replace V g (), R, and with the Thevenin equivalent, we no longer have a point in the circuit where I() i flowing. Neverthele, we may till compute I() from V(): I() V() C R (Note that V i acro R and /C a well a acro L.) Our circuit model become: R R2 V g () R R R2 V() - C - L The impedance of the L and C in parallel i L C L C C L L 2 C 2
Ue V-divider to find V: V() V g () R R Now implify the expreion. V() V g () R R L C C L R C 2 C 2 Multiply top and bottom by 2 /: R R V() V g () C R C 2 Rewrite the voltage divider a R / : V() V g () R Divide top and bottom by R : C C 2 R V() V g () 2 C R 2 R R Tidy up the denominator and ubtitute V g /: V() 2 R 2 V ha the form of a decaying ine wave: R
L{ ke αt in(ωt) } where k i a real contant and kω ( α) 2 ω 2 kω 2 2α α 2 ω 2 kω, α2 ω 2, and 2α R. ASIDE: The fundamental form of the polynomial appearing in V() i independent of the value of R,, L, or C. If I() i to be 90 out of phae with V(), we mut hope that it can be made to have the form of a decaying coine: L{ k 2 e αt co(ωt) } k 2 ( α) ( α) 2 ω 2 k 2 ( α) 2 2α α 2 ω 2 where k 2 i a real contant (poitive or negative). I() i V() divided by the parallel impedance of R and C: I() V() C R The parallel impedance of R and C i R C R C R C Subtituting for V() give I() 2 I() 2 R R C. R R R C R C R R 2 Above, we concluded that I() mut be of form R R C R
I() k 2 ( α) 2 2α α 2 ω 2 The denominator polynomial in i in the correct form. Matching the numerator give: k 2 ( α) R C We mut have k 2 / and α /R C. Earlier, we found α ( )C. 2 R We conclude that ( ). R 2 R The olution to thi equation i R 500 Ω, (value given in problem tatement): R 500Ω A tandard value for R would be 50 Ω. Now we find C. We look for an equation with C a the only unknown. The problem tate that /α T or α /T ω/2π: α ( )C 2π ω 2 R From the contant term in denominator of V() and I(), a noted earlier, we have an equation involving α and ω: α 2 ω 2 Subtituting for α and ω, we have ( )C 2 R 2 [ ( 2π) 2 ] Inverting both ide, we have. [ 2( R )C] 2 ( 2π) 2. Rearranging and canceling one power of C on both ide give a value for C:
C ( 2π )2 ( ) 2 L 4 R ( ) 2 ( ) 2 200mH 2π C 4 500Ω 500Ω ( 2π) 2 C 4 250 250Ω [ ] 200 C ( 2π) 2 C 32.4µF 250 ( 2π) 2 200mH 2 k 250Ω 2 200mH µh Ω 2 A tandard value for C would be 33 µf. ol'n: (b) From (a) we know v(t) ke αt in(ωt) and i(t) k 2 e αt co(ωt). From (a) we alo know k ω 2π α 2π 2π and k 2. k V 59 mv 2π k 2 V 500Ω 2 ma Note: We ue V in the numerator becaue V g (t) actually tranform to V/ rather than /. We leave out the V to avoid clutter and confuion in the calculation. We might mitakenly think of the V a V(). For α and ω, we have α 500Ω 32.4 µf 6.7/ and ω 2πα 388rad/. Subtituting numerical value for k, k 2, α, and ω in the time-domain form for i(t) and v(t) give our final anwer.
v(t) 59 e -6.7t in(388t) mv i(t) 2 e -6.7t co(388t) ma Conitency check: i(t) i R i C v(t) C dv(t) k e αt in(ωt) C d R dt R dt [ke αt in(ωt)] k e αt in(ωt) Ck [( α)e αt in(ωt) e αt co(ωt) ] R k[ ( α C C R RC e αt )] in(ωt ) (Ck ) RCω ω e αt co(ωt) R e αt co(ωt) i(t) (ince R )