Question

We create a new circuit that consists of a single loop with al Farad capacitor, current probe, and single light bulb in series with one another. We add two smaller loops by attaching a voltage probe to measure the voltage across the capacitor and by attaching the leads from a hand-crank generator across the light bulb. We then remove the light bulb from the circuit and start a data collection. We use the hand-crank generator to charge the capacitor, then disconnect the hand-crank generator and soon afterwards reinsert the bulb into the circuit to complete the large loop. We allow the capacitor to fully discharge through the light bulb before stopping the data collection. We then use the mouse cursor to find the starting voltage of the capacitor when the light bulb was plugged in. What was the voltage across the capacitor at the moment when the light bulb was plugged in? 4.8 V How much stored energy does this represent? 1.0 Integral for Latest Power Integral 8.299 W's 0.5 Power (W) 0.0 -0.5 -1.0 0 (69.76, 0.843) 50 100 150 Time (s) According to the plot, how much energy was dissipated in the light bulb? What percentage of the energy that was stored in the capacitor was not dissipated in the bulb? What happened to this energy that was not dissipated in the light bulb? In other words, where did it go when it left the capacitor?
1.5 1.0 Potential 0.5 Auto Ft for Latest Potential Pot Apexpl-C+B A 1481 9.873E-05 C 0.01506 - 1.973E-06 B 0.003797 +3.455E-05 Comelation: 1.000 RMSE 0 002510 V 0.0 100 300 0 (180.9, 0.809) 200 Time (s) According to the curve fit parameters, what was the time constant during the early part of the discharge? According to the curve fit parameters, what was the time constant during the later part of the discharge? Why did the time constant change? Your answer should include a description of what physical property in the circuit changed and why this would affect the time constant​

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