Irodov – Problems in General Physics

4.112. Q= (^17 ) 17 -^16 = = 1 .0 10^2.

4.113. (P) =R (I 2 ) 112 RA = 20 mW.

4.114. (P) = 1 i 2 RCV,%/L = 5 mW.

4.115. w= /- V LC 4R 2 C 2 ' R <- /-.

4.116.-A-- — and — — = L, L2 L R 2 — R •

4.117. I= V° te-t/V7-7c. ' I—Imax 1 —17. e c- at the moment

t,n =1/17.

4.118. I—

Vrn

[cos (cot — cp)—cos cp • e-tR/L], tan cp = coL/R.

iiR2H-co 2 L 2

4.119. I=

vm

[cos (wt cos cp • e-t/RC], tan cp =

1R2+1/((0C) 2

coRC -

4.120. The current lags behind (^1)
UcR
the voltage by phase angle cp, defined (^) Axis (^) Axis I
—L
by the equation tan cp = 1A°^4 nnp^2 'va^ SRO
of current (^) I of current
4.121. The current is ahead of^
(a)^ r
the voltage by the phase angle (^) (b)
= 60°, defined by the equation Fig. 31.
tan cp = (17,„/R/ni)^2 —1.
4.122. (a) V' =Vo-i-V,,cos(cot— a), where Vm=1^70 /V-1 (coRC) 2 t
= arctan (WIC); (b) RC =V 1 2 -1/co= 22 ms.
4.123. See Fig. 31.
4.124. (a) = 17,n/YR^2 + (coL —1/coC)^2 = 4.5 A; (b) tan cp
,cp = — 60° (the current is ahead of the voltage);
(c) V c --= /,,,/coC= 0.65 kV, VL = 4,11R 2 + co^2 L^2 0.50 kV.
4.125. (a) co = ci4 — 2f3 2 ; (b) w -= coVV co^2 ,, — 2P^2 , where co: =
= 1/LC, 13 = RI2L.
4.126. For C = (^) = 28 iiF; V L = Vin (coL/R) 2 = 0.54 kV;
Vc = V nicoLIR= 0.51 kV.
vrn
4.127. I =In, cos (cot + cp), where R (coRC) 2 and
tan cp = coRC.
4.128. coo = c (L1L2L2 _ 14.2)^
4.129. Q=V12 2 — 1/4.
rig-1 1
4.130. Q= 17 (n— 1) 2 4 •
4.131. (a) coo = V 6)0) 2 ; (b) Q=1/
0 )1( 0 2 (n^2 —^1
((0^2 —04)^2 4 •
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