Question

4. Suppose that you are supplied with this equipment = 0.012 2), plus 2 switches and plenty of wire (of negligible resistance) 1 non-ideal battery (with internal resistance 1 parallel-plate capacitor, initially uncharged 4 identical ohmic resistors, each with resistance R 4.00 Q 2 solenoid inductors, with inductance values L and L, respectively. Inductor A is half as long as inductor B Inductor A has a diameter of 5.30 cm; inductor B has a diameter of 6.36 cm. r And you conduct the following experiments Experiment (i): Using the two inductors as the coils, you build a simple transformer Then you connect coil A across an AC power source; that is, coil A is the primary coil Result: The (AC) voltage difference across coil B (the secondary coil) is 1200 V. Then you reverse the transformer-now it's coil B as the primary coil (connected across the same AC power source). Result: The (AC) voltage difference across coil A (now the secondary coil) is 48 V Experiment (i): You build the circuit shown here (with the switch initially open). Then you close the switch at t 0. Result The capacitor has 75.0% of its full charge at = 11.124 ms R You leave the switch closed for a total of 48.958 s, then open it and dismantle the circuit, setting the charged capacitor aside for later. switch A switch B Experiment (ii): You build the circuit shown here. Then you close both switches and leave them closed for a long time R R Results: The energy stored in inductor B is 1.7795 J; and the total power being produced by the two resistors on side B is 632.72 W L Continuing from experiment (iii), now you open switch A at t = 0 Calculate the total power being produced by the two resistors on side A at а. 0.389 ms. b. If you were to dismantle the above circuit and instead simply connect inductor B and the charged capacitor as shown here, then close the switch at 0, what percentage of the total energy stored in the circuit will be in the electric field at 9.26 ms? C L

Answer 1

LI. We knos hat Nno tu^ns A area D2 (danu) ongth L 2. LA NA NB. 5:30 6.36 18 LA 25NA Oh perimnt Primasy Ceil B as CeriA Phimasy Cas Vi NA NA hom e 3 1

fxporimed Charge Aelationapacikr giuen Cta.(-e) vis (ArR)C inon at t: 11-124 ms, -= O.75 -124 6.75 OY hene T 8.046 m F ntR y.ol2 at t: 8.958 I- e Uollag E at t-urgsis changed Laith Wnce Capacit is perinent SR atter long SR SR R 2 R TE tsteedy stah, inductor basruven act a shott cht il egually duidk in fr Renishr R any E E 4.o48

Energytodn indvetw B 2 58.319 CY JbEt'] 2 4.048, Lieas total eneng producud by tuso ausisloms en side B 2 632.72 (4.o48, Oh E = 36.085 Vol iing h n e 4.727 mH fhom eO LA = u4.787 mAH LA = 3.4.88 mAH (a) Sutch openud at to sínce cunount throngh indvetr chan Audeldy 8.914 A ZR GoT Can not LA E (t-o) Cuvunt ea fr indrlur i n LA as 2R at t: 0.3 ?9 ms 8.914 e : 2.5r2 A A pradurd by tso uni stong = 2R- 2d2s51).y total powr

b) tor Lcircuit, cwnnt enalio vi as R ushure T IR at t o yo48 -a Sint Vostage avros CLdz at t 9.26 ms u4.787 9.26 Sin Tчизет. TT 4.0 C* £&t*hh 35vol I Seckic fieldcv. Energy stored as 2 .225J -3 Tuilicstored enoy; Ч-о43. 2 6tt Percetage engy stored as elechric fiald at 1= 9.26ms 68.8 779