Survey

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Survey

Document related concepts

no text concepts found

Transcript

4/27/22, 10:15 PM Powered by Homework Hub Powered by Homework Hub | J.D#1467 || Dev | hello#1146 ANSWER GIVEN BY [Anonymous] Jaffa_the_warrior's server likes 1 dislikes 0 QUESTION file:///C:/Users/user/Downloads/Answer (7).html 1/7 4/27/22, 10:15 PM Powered by Homework Hub (b) In an electrical circuit shown in Figure 2.1 below, when a capacitor, C, is being charged through a resistance, R, by a battery which supplies a constant voltage, Eſt), the instantaneous charge, Q, on the capacitor satisfies the differential equation: R499 = E(1) dt C S(Switch) 강 с Figure 2.1 Electrical Circuit Given that the generator having an file:///C:/Users/user/Downloads/Answer (7).html 2/7 4/27/22, 10:15 PM Powered by Homework Hub electromotive force of E(t)=120sint Volt is connected in series with a 6 Ohm resistor and a capacitor of 3 Farad. The switch is closed at time i=0 with a step size of h=0.5. i. Find the approximated value of the instantaneous charge, Q, on the capacitor, from 1 = 0 to 1 =1.0 by using Ralston's second order Runge- Kutta Method. Perform the calculation up to four decimal places. (13 marks) . ii. If someone repeats the computation with h=0.55 to approximate (1.0), will the approximation value be closer to the actual solution? Justify your answer. (2 marks) Show transcribed image text (b) In an electrical circuit shown in Figure 2.1 below, when a capacitor, C, is being charged through a resistance, R, by a battery which supplies a constant voltage, Eſt), the instantaneous charge, Q, on the capacitor satisfies the differential equation: R499 = E(1) dt C S(Switch) 강 с Figure 2.1 Electrical Circuit Given that the generator having an electromotive force of E(t)=120sint Volt is connected in series with a 6 Ohm resistor and a capacitor of 3 Farad. The switch is closed at time i=0 with a step size of h=0.5. i. Find the approximated value of the instantaneous charge, Q, on the capacitor, from 1 = 0 to 1 =1.0 by using Ralston's second order Runge- Kutta Method. Perform the calculation up to four decimal places. (13 marks) . ii. If someone repeats the computation with h=0.55 to approximate (1.0), will the approximation value be closer to the actual solution? Justify your answer. (2 marks) ANSWER file:///C:/Users/user/Downloads/Answer (7).html 3/7 4/27/22, 10:15 PM file:///C:/Users/user/Downloads/Answer (7).html Powered by Homework Hub 4/7 4/27/22, 10:15 PM file:///C:/Users/user/Downloads/Answer (7).html Powered by Homework Hub 5/7 4/27/22, 10:15 PM Powered by Homework Hub i hope i clearly answered your question .plz kindly give me an upvotes for my effort for doing this. I hope you understand. Thanks have a nice day🙂 b) @ on the capacitor is given by The instantaneous charge the differential equation Role + C = E(+) Elt R do + с where R=bu, 6 c = 3 F Elt)=120 sint v switch is closed at t=o. & max = Emexic = 19013 Qo=&max = 360 = Q 6 da + a 3 120 sint gli 81 da 20 sint Q 18 (1) JI t At =0, Q = Q. 8-9 Qo - 360 Take step size of h=0.5 9) R-K Using 2nd order Method. for dy fix,y). Subjected to condition at x=do, y=y. da We have to find the value of y find the value of y at x = ko+AX Suppose at x=notan, y = yo tay x 8 + let Dx=h, by = kitkz 2 where k, = hflxo, yo) K2 = hf (26th, Yo tki) . B = ) So Applying R-k and order Method for eqn (1) = 20 sinta do 18 To find Q (1) = 2 to=0, 0 (0) - Qo, h = 0.5, = Q , t,=0.5, Q1 = @ot sy kitk2 Ay = 10 18 ki - htlto, Ro) 0.5 (20 sinto - 360 ) h+10.5, 80+k) = ht10.5, 350) (Qosinco.s)- 350 2 ) 0.5 18 K2 = -4.9279 Kitke -10 -4.9279 by = =-7.4639 2 2 - Qi= do ty, 360 - 7.4639 Q1 = 3525361 ty = 0.5 Now Q+ 2 / -, , KFla ta=1 Q₂ = Q, toy, Dy = kitkz kahf (1,Q) - hf 10.5, 359.5341) = , 0.5 frosin (0.5) - 359.536) K2= hf (tith, Qi+k) 0-5 (Qosinli) - 347-5372] = {( = -4.9984 18 ] 0.54 (1, 347.5377) , -). 239 -4.9984-12391 - 3:11875 by, = Kitka 2 Az Q2 = &2 = Qit Oy, = toy 352.536) - 3:1187 = 3५१.५।740 The approximate value of the instantaneous charge & the capacitor at tal, Q(1) = 349.4174 (1 us on ii) of h=0.55, to -0, file:///C:/Users/user/Downloads/Answer (7).html 6/7 4/27/22, 10:15 PM Powered by Homework Hub = Q. = 360. 5,= 0.55 Q, = Q. + lo+DY = Kitka 360) = -14 . - By ki = hf 1to, Q. ) = 0.55410,360) 0.5510- K2= ht(toth, 2o+k)= 0.5 +10.55, 349) Qo 0. ssl ao sin(055) - 349) = -4.9143 ; -11-49143 by : -7.9571 2. ti = 0.55, 1 Qi 360 -7.9541 Qi = 352.0429. . 1 to = H, +0.55 = lol Q2 = 0, +ay - Dit By DY ki+k2 Q 5. 0073 11 18 ki ht(t,,Q,) = 0.55[ do sin 10.55) - 352.0429 ) 20 ) kzht,th, Q,+ki) = 0.557(9-), 347.0356) ( =0:55 [ao sinli. 1) - 347.0357 342-035] = -0.8806 DY -5-0073-0.8006 = -2.9039 2 349.139 82 - 352.0429-2.9039 = file:///C:/Users/user/Downloads/Answer (7).html 7/7