Question

4. Suppose that you are supplied with this equipment: 1 non-ideal battery (with internal resistance r 0.012 Q), plus 2 switches and plenty of wire (of negligible resistance) parallel-plate capacitor, initially uncharged. 4 identical ohmic resistors, each with resistance R 4.00 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 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 (ii): 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 t 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 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 0 Calculate the total power being produced by the two resistors on side A at t0.389 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 0, what percentage of the total energy stored in the circuit will be in the electric field at / = 9.26 ms?

Answer 1

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