Second order linear differential equation representing momentary charge in the given RLC series... Cosine varies between -1 and 1. cos[√(1/LC-(R/2L)²)t]=1 when t=0 or when:

1/LC-(R/2L)²=0, 1/LC=R²/4L², R²=4L/C; and cos[√(1/LC-(R/2L)²)t]=-1 when:

√(1/LC-(R/2L)²)t=π. Therefore R²=4×50000=200000, R=447.21 ohms. R cannot exceed this value or the square root could not be evaluated. R/2L≤447.21/10=44.721.

When R=0, q=q₀cos(t√2000); when R=447.21, q=q₀e^-44.721t.

When q/q₀=0.01=e^-44.721t, -4.6052=-44.721t, t=0.3256s.

If t=0.05, 0.01=e^(-0.005R)cos[0.05√(2000-0.01R²)].

Consider f(R)=e^(-0.005R)cos[0.05√(2000-0.01R²)]-0.01. We need two values of R (R₁ and R₂) such that f(R₁)<0 and f(R₂)>0. Since R≤447.21, let’s try R₁=300 and R₂=400.

f(300)=-0.0295 and f(400)=0.0631. So we know that 300<R<400.

The table below shows a range of values of R between 300 and 400:

R      f(R)

300  -0.0295

320  -0.0082

325  -0.0032

330    0.0018

There is a change of sign between 325 and 330, so 325<R<330.

R      f(R)

325  -0.0032

326  -0.0021

327  -0.0011

328  -0.0002

329    0.0008

328<R<329

R          f(R)

328.0 -0.00015

328.2   0.00005

R                f(R)

328.00 -0.0001507

328.15 -0.0000014

328.16   0.0000085

R                       f(R)

328.150 -0.000001422

328.151  -0.000000427

328.152   0.000000568

R                   f(R)

328.1510 -0.000000427

328.1512 -0.000000228

328.1513 -0.000000128

328.1514 -0.000000029

328.1515   0.000000071

R                    f(R)

328.15140 -0.00000002894

328.15142 -0.00000000904

328.15143   0.00000000091

So R=328.15143 to 5 decimal place accuracy.

Now compare that with the Regula-Falsi Method:

We can work out the equation of the line that joins the two points (300,-0.02950255687) and (400, 0.0631219656):

Slope=(0.0631219656+0.02950255687)/100=0.00092624522.

Line is y-0.0631219656=0.00092624522(x-400). When y=0, 0.00092624522(x-400)=-0.0631219656, x=400-0.0631219656/0.00092624522=331.8517776. So R₁=331.8517776.

We know from the above method that f(R₁)>0.

Therefore 300<R<R₁.

f(R₁)=0.00365859728, f(300)=-0.02950255687.

Slope between the points is (0.00365859728+0.02950255687)/31.8517776=0.0010410843.

Line is y+0.02950255687=0.0010410843(x-300), so x=300+0.02950255687/0.0010410843=328.3382977. R₂=328.3382977.

f(R₂)=0.00018593343, f(300)=-0.02950255687.

Line is

y+0.02950255687=(0.00018593343+0.02950255687)(x-300)/28.3382977=

0.0010476455(x-300).

x=300+0.02950255687/0.0010476455=328.1608205. R₃=328.1608205.

f(R₃)=0.0000093474, f(300)=-0.02950255687.

Line is

y+0.02950255687=(0.0000093474+0.02950255687)(x-300)/28.1608205=

0.00104797743(x-300).

x=300+0.02950255687/0.00104797743=328.15190102.

R₄=328.15190102.

f(R₄)=0.00000046973.

y+0.02950255687=(0.00000046973+0.02950255687)(x-300)/28.15190102=

0.00104799411(x-300).

x=300+0.02950255687/0.00104799411=328.1514528.

R₅=328.1514528.

R₆=328.15143 to 5 decimal places. Compare with 328.15143 obtained earlier.

by Top Rated User (884k points)

Unable to tabulate or upload photo of table because of persistent system fault. I hope you can follow the logic and create a table for yourself.