4.90. (a) Q 1/2 17(0 4 0782 )a S tan2 1 — 2.2; (b) A = nma2 (4- (02) tan q ------ 6 mJ. Here coo = V x/m.
4.91. (a) (1))- - - -
F8136)21'n
• (b) co= (00, (P)max= FO/4Pm•
(cos— 0)2+4020
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/q),./N.)2— 1
2 sin a
4.94. co —17ne2/som = 1.65 .1016 s -1.
4.95. V2 + /2L/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 = (12 - 1) W.
4.98. (a) T = 231 V L (C C2) =- 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 ( — (5
7 )± 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/CR2 =-- 5.
4.105. L= L1 -h L2, R= R1 ± R24.106. t=
ln 1 = 0.5 s.
4.107. n = 21 31 174 1 ---- 16.
4.108.
0)0
— 1
0)0
17 + /(202
8Q2 = 0
.5%.
4.109. (a) Wo = 1/262 (L
CR2)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 current in the coil winding, I' is the leakage current (I' = V I R).