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```4/27/22, 10:15 PM
[Anonymous]
Jaffa_the_warrior's server
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QUESTION
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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
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
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doing this. I hope you understand.
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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,