Question

Suppose that you are supplied with this equipment 4. Inon-ideal battery (with internal resistance r 0.012 ), plus 2 switches and plenty of wire (of negligible resistance). I parallel-plate capacitor, initially uncharged. 4 identical ohmic resistors, each with resistance R 400 Q 2 solenoid inductors, with inductance values L and Ly, 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 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 = 0. Result: The capacitor has 75.0% of its full charge at t 11.124 ms You leave the switch closed for a total of 48.958 s, then open it and dismantle the cireuit, setting the charged capacitor aside for later switch A switch B Experiment (ii): You build the circuit shown here. Then you L 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. R 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 t 0.389 ms a. 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 t0, what percentage of the total energy stored in the circuit will be in the electric field at / = 9.26 ms? C

Answer 1

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