Question

X resistances in the circuit can be neglected. The switch is closed at time t-0 X X 1 X E- What is the acceleration of the bar as a function of a) time? b) What is the velocity of the bar as a function of time? Figure 3-for Problem 29.60 Chapter 30 Problems: Problem 5 - variation of problem 30.64: Using the circuit shown in Figure 4, a) Find the reading in each ammeter and voltmeter just after switch S is closed b) Find the reading in each ammeter and voltmeter after S has been closed a very long time. c) At what time does the power in the inductor equal S the power in the 75 Ω resistor? 100.02 15.0mH Problem 6- variation 30.74: In the circuit shown in Figure 5, neither the battery nor the inductors have any resistance, the capacitors are initially uncharged, and the switch S has been in position 1 for a very long time. a) What is the current in the circuit? The switch is now suddenly flipped to position 2. b) Suppose R-0, find the maximum charge that each capacitor will receive, and how much time after Figure 4 - for Problem 30.64 the switch is flipped it will take them to acquire this charge. c) Find the value of R to make the LRC circuit critically damped. 50.0V 50.0 75.0Ω 1 S 2

i need help with problem 5

Answer 1

100 Ω so v 50 2 45요 cive is debexmina, just after closing the Soitok keadings of ammeten and voltmeter. Juat afte the suwitch is closed ndacoywil provide a inPinite ust a resistance, that is eux-tent in induct。γ branch uilt be 旡ere V, vo ts. V abye o volts ciscuit wil b e com es Aften the Soitch S is closed the as rheon belew

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ne -3 し(ニ0.333x50 Vq- 16.65 volts readinge after Now se cal culate voltmeter and oummoter aften switch S has o lora been closed a v Afber a Suitsh is closed Por ong ime inductoy coill provide 굿ere Yeststances that inductor coos k líka a conducting utYe "The circuit aill become Pty 00n 35Ω At vex tots 50,(2 and 75 f2 axe connected in n p. S oCaneplace both, by 50 75 6095 3+50 125 And neuo R-30 요. and 100 are cennectedn seues . So equivatont vesistance of e circuit is e30 Io

ne e 30 From ohms lauo 50 Re 130 L20.384 amps kk1 cho Picar ent law ,in the circuit shoon above Bausing 0.384 1,+ Thon 50 and ane Connecbed in paneSpotent ial diffexence acoss bthill be same. ie 1 50 7-5 0.666 Substibut 1, value in ohove 0.666 T0384 1.66610.384 10.38 ,666 12-0, 2.304 Amps Ihun 1, will be 0.666 x 02304 , 01535

Vi-38.4 volts V0 volts 5 volts V 1x 50 0, 2.304-X 50 A, 스 0-384 Amps 3 A3 0.230 C)we cletenmine ot ohat tlme does tha powen in the indactor Abang time (t) the current inkhe inductor is; t dj

L-15 m H and 二0.1535 - 5000 t ono power in the inductor is that tha eauation and power in 75 ΩγΕΝ storis, tory@ and ④ Equate tha dt 000t)| © lo.is35 (t←e 15 01535 (1-b 75 -5000七 500 ob -5000t - 5000 t 5000 lyvahar ol both sides

f 5000 t, -メ0-69 3 t 0.693 5000