The voltage at the terminals of the 0.6 µF capacitor shown is 0 for t < 0 and 40e-15000t sin 30000t V for t > 0. Find a) the current i(0); b) the power delivered to the capacitor at t = t/80 ms; c) the energy stored in the capacitor at t = t/80 ms. C +
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- Please draw the problem and answer, thank you. A capacitor is made of 2 rectangular metal plates with side length of 3 cm x 6 cm separated by a distance of 2.36 cm with water in between the plates. The capacitor has a voltage of 110 v and is not connected to a battery. Calculate the capacitance. What is the new capacitance if we replace water with a new dielectric material with a constant of 3.75 in between the plates? What is the new voltage? What is the charge on each plate?At 0-, no currrent flows through the capacitors because they are open, how did you combine the capacitors for the voltage divider since it is the capacitance value and not reactance, is it right? Can we just combine the capacitance value? Please explain. I did not understand..To measure the capacitance C of a capacitor, you attach the capacitor to a battery and wait until it is fully charged. You then disconnect the capacitor from the battery and let it discharge through a resistor of resistance RR. You measure the time T1/2 that it takes the voltage across the resistor to decrease to half its initial value at the instant that the connection to the capacitor is first completed. You repeat this for several different resistors. You plot the data as T1/2 versus R and find that they lie close to a straight line that has slope 9.00 μF. What is the capacitance C of the capacitor?