Irodov – Problems in General Physics

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4.133. / 0 // = -V1 (Q 2 + 1/ 4 ) (1 2 — 1) 2 /1^2 , 2.2 and 19 respectively.
4.134. t = i/zato.
4.135. (a) I = —Io z 1.154; (b) I = Y8 Io 1.1110-
R
4.136. 2/tL, 1= 2 kHz.
4.137. The current lags behind the voltage by the phase angle

(IDarccos P= -;1/^1 Z^2 - Xi = 0.16 kW.
172
4.138. For R= 0.20 k52; P „ „ = 20)L --= 0.11 kW.


4.139. Increased by 30%.
4.140. For Q >> 1 the^ ratio^ is^ A0)/(60 1 /21/ — 1/Q = 0.5%.

IR Axis of

Ic

Axis of
IR Axis of

voltages volt ltages g es (^) voltages
\ 10
\ I (^) -- IO
IL Iv (^) I Lx IL
(a) (6) (^) (c)
Fig. 32.
4.141. P = 2 112 (V 2 — 1/ — VDIR = 30 W.
4.142. P 1 = 1 / 2 (1 2 — — R 2.5 W.
4.143. Z = R/1/1 (o)CR) 2 = 40 O.
4.144. See Fig. 32.
4.145. (a) w,es


LC 3 .10^4 rad/s; (b) / =
VRC/L =-.
=3 mA, / L = V VCIL = 1.0 A, /c = V 17TC: —^
) 2 =
1.0 A.
4.146. tan cp —
(t)C (R2+ co2L2)— "



  • 4.147. Z=


R 2 ± cu 2 L 2
(0)cR)2+ (1— w 2 CL) 2 •

oi (^2) L 2 Li 2 /3 8L 12
4.149. (Fx)=2 (R 2 + OLD ox •
2/
4.150. t =
a (VT+ liT2)
4.151. Acp =1 I (x 1 — x 2 ) cos a + Y2) cos 0+(z1-z2) cos
4.152. k=w v 1 V2 V3
4.153. t = a cos [(1 — V/v) cot — kx^1 ], where v = (o/k.
4.155. (a) a/X, = 5.1 .10- 5 ; (b) v,„ = 11 cm/s, 3.2.10 -4;


(c) (at/ax),,, = 3.2.10 -4, (avat),,,---1, (alai),,,, where v = 0.34 km/s


is the velocity of the wave.

330
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