4.90. (a) Q 1 / 2 17 (0 40782
)a Stan2 1 — 2.2; (b) A = nma 2 (4-
- (0 2 ) tan q ------ 6 mJ. Here coo = V x/m.
4.91. (a) (1))---(cos— 0)2+4020 - (^) F8136)21'n^ • (b) co= ( 0 0, (P)max= FO/ 4 Pm•
4.92. (P)max—(1)). 100 s.^
(P)max 12 -^1
4.43. (a) A = — apmN,, sin a; (b) Q — 1/(cos a ± 2o)2 sin^2 /q),./N.)a^2 —^1
4.94. co —17ne 2 /som = 1.65 .10 16 s - 1.
4.95. V 2 + / 2 L/C = V.
4.96. (a) I = In, sin coot, where / 7 „ = V,,VC/L, coo = 1/1/ LC;
(b) gs= Vm/Ya
4.97. A = (1 2 - 1) W.
4.98. (a) T = (^231) V L (C C 2 ) =- 0.7 ms;
(b) in, ---- (^) + COIL = 8 A.
4.99. V = 112 (1 ± cos cot) Vo, where the plus sign refers to
the left-hand capacitor, and the minus sign to the right-hand one;
4.100. I =-1- 1. cos (07 LC).
4.101. (a) 4, =^ ; (b) to = 4-) [arctan (—^7 (^5 )± gm]. Here
n = 0, 1, 2, ...
4.102. Vo/Vn, -17 1—
R42L
4.103. Vc = .1.7n V L/C e-fit sin (cat + a) with tan a = co/f3; Vc (0) =
L
= /,„ V C (1+ PAO) •
4.104. WL/ We = L/CR 2 =-- 5.
4.105. L= L 1 -h L 2 , R= R1 ± R2-
4.106. t= ln 1 = 0.5 s.
4.107. n = 21 174 1 ---- 16. 31
4.108.
0)0
0)0 — 1
17 + /(202 8Q2 = 0
.5%.
4.109. (a) Wo = 1 / 262 (L CR^2 )I(r R)^2 = 2.0 mJ; (b) W =
= Woe-tR/L = 0.10 mJ.
4.110. tx In =1.0 ms.
4.111. (a) al ---- 17 - A- 4R2c2 ; (b) Q = V -^4 1.
When solving the problem, it should be taken into account that
dq/dt = I — I', where q is the charge of the capacitor, I is the cur-
rent in the coil winding, I' is the leakage current (I' = V I R).