3. Расчётные формулы и вычисление электрических величин
3.1 Расчёты домашнего задания:
XL= ω•L= 314•0.1= 31.4 (Oм);
1
ω•С
XL
C
1
314•6•10-5
XС = = ≈ 53.08 (Oм);
Z=√R2+ (XL─XC)2 = √ 900+ (53,08─ 31.4)2 ≈ 37,01;
XL─XC
R 31.4─53.08
30
φ=arctg == =≈ -35.85;
ωРН=1/√L•C =1/√0.1 •60 •10-6 ≈ 408.25 (Гц);
f=1/ (2•π•√L•C) ≈ 64.97 (Гц);
I= U/Z=10/ 37.01 ≈ 0.27 (A);
UR=R• I = 30 • 0.27≈ 8.11 (B);
UL=XL• I=31.4 • 0.27≈ 8.48 (B);
UC=XC • I=53.08 • 0.27 ≈ 14.34 (B);
UK=√UR2+UL2 =√R2+XL2 •I = √900+31. 42 • 0.27 ≈ 11.73 (B);
P= R• I2 = 30 • (0.27)2 = 2.19 (Bт);
Найдём резонансное сопротивление:
CPH =1/ (ω2 • L) = 1/ (3142 • 0.1) ≈ 101.42 мкФ.
U
Расчёты цепи.
1). UR = I•R = 0.04•30 = 1.2 (B);
UL = XL•I = 31.4•0.04 = 1.26 (B);
XL = ω•L = 314•0.1 = 31.4 (Oм);
XC = 1/ (ω•C) = 1/ (314•10-5) ≈ 318.47 (Oм);
Z = √ R2 + ( XL─ XC)2 = √ 900 + (318,47 ─31.4)2 ≈ 288.63 (Oм);
XL─
XC
R
φ = arctg = arctg [(31.4─318.47)/30] ≈ -84.03 (град);
Z = √ R2 + ( XL─ XC)2 = √ 900 + (318,47 ─31.4)2 ≈ 288.63 (Oм);
2). UR = I•R = 0.18•30 = 5.4 (B);
UL = XL•I = 31.4•0.18 ≈ 5.65 (B);
XL = ω•L = 314•0.1 = 31.4 (Oм);
XC = 1/ (ω•C) = 1/ (314•10-5•3) ≈ 106.16 (Oм);
Z = √ R2 + ( XL─ XC)2 = √ 900 + (106.16 ─31.4)2 ≈ 80.56 (Oм);
XL─ XC
φ = arctg = = arctg [(31.4─106.16)/30] ≈ -68.14 (град);
R
3). UR = I•R = 0.48•30 = 14.4 (B);
UL = XL•I = 31.4•0.48 ≈ 15.07 (B);
XL = ω•L = 314•0.1 = 31.4 (Oм);
XC = 1/ (ω•C) = 1/ (314•10-5•5) ≈ 63.69 (Oм);
Z = √ R2 + ( XL─ XC)2 = √ 900 + (63.69 ─ 31.4)2 ≈ 44.08 (Oм);
XL─ XC
φ = arctg = = arctg [(31.4─63.69)/30] ≈ -47.11 (град);
R
4). UR = I • R = 0.7 • 30 = 21 (B);
UL = XL • I = 31.4 • 0.7 ≈ 21.98 (B);
XL = ω • L = 314 • 0.1 = 31.4 (Oм);
XC = 1/ (ω • C) = 1/ (314 • 10-5 • 7) ≈ 45.5 (Oм);
Z = √ R2 + ( XL─ XC)2 = √ 900 + (45.5 ─ 31.4)2 ≈ 44.08 (Oм);
XL─ XC
φ = arctg = = arctg [(31.4─ 45.5)/30] ≈ -25.17 (град);
R
5.) UR = I • R = 0.36 • 30 = 10.8 (B);
UL = XL • I = 31.4 • 0.36 ≈ 11.30 (B);
XL = ω • L = 314 • 0.1 = 31.4 (Oм);
XC = 1/ (ω • C) = 1/ (314 • 10-5 • 9) ≈ 35.39 (Oм);
Z = √ R2 + ( XL─ XC)2 = √ 900 + (35.39 ─ 31.4)2 ≈ 30.26 (Oм);
XL─ XC
φ = arctg = = arctg [(31.4─ 35.39)/30] ≈ -7.58 (град);
R
6). UR = I • R = 0.32 • 30 = 9.6 (B);
UL = XL • I = 31.4 • 0.32 ≈ 10.05 (B);
XL = ω • L = 314 • 0.1 = 31.4 (Oм);
XC = 1/ (ω • C) = 1/ (314 • 10-5 • 11) ≈ 28.95 (Oм);
Z = √ R2 + ( XL─ XC)2 = √ 900 + (28.95 ─ 31.4)2 ≈30.10 (Oм);
XL─ XC
φ = arctg = = arctg [(31.4─ 28.95)/30] ≈ +4.67 (град);
R
7.)
UR = I • R = 0.30 • 30 = 9.0 (B);
UL = XL • I = 31.4 • 0.30 ≈ 9.42 (B);
XL = ω • L = 314 • 0.1 = 31.4 (Oм);
XC = 1/ (ω • C) = 1/ (314 • 10-5 • 13) ≈ 24.50 (Oм);
Z = √ R2 + ( XL─ XC)2 = √ 900 + (24.50 ─ 31.4)2 ≈30.78 (Oм);
XL─ XC
φ = arctg = = arctg [(31.4─ 24.50)/30] ≈ +12.95 (град);
R
Добротность контура равна:
(XL)PH = ω • L =314 • 0.1 = 31.4 .
Q = 31.4 / 30 = 1.05.
Построим графики зависимостей XL (С), UL(С), XC(С), φ(С), UC(С), I(C):
Вывод:
Отчет принял преподаватель:
(Ф. И. О.)
" " 200 г.
подпись преподавателя.
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