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(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
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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
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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,
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= 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 =
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