We consider the discharge process of a parallel plate capacitor of Capacitance C, through a resistor of resistance R. C

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We Consider The Discharge Process Of A Parallel Plate Capacitor Of Capacitance C Through A Resistor Of Resistance R C 1 (63.66 KiB) Viewed 67 times

We consider the discharge process of a parallel plate capacitor of Capacitance C, through a resistor of resistance R. C is defined as ususal, as C=q(t)/(t); note that no matter what the numerator and the denominator over here are time dependent; C remains constant throughout; q(t), is the charge instensity at either plate at time t; its value at t=0 is then q0); V(t) is the electrci potential difference between the plates of the capacitor at hand at time t; its value at t=0, is then V(0) a) Sketch the circuit. Write the differential equation describing the discharge. Show that q(t)=(O)expft/RC), thus, i(t)=i(O)exp(- t/RC). Express i(0) in terms of VC) and R. Note that here, you should write i(t)-dq(t)/dt. Why? Sketch, V(t), i(t) ve q(t), with respect to t. b) As the capacitor gets discharged, it throws its energy through R. The enery discharged per unit time is by definition de/dt; this is, on the other hand, given by Rif(t). Show then that, the total energy E thrown at R, as the capacitor gets discharged, is (1/2)CV²0). (Note that this is after all, the “potential energy" stored in the capacitor.) c) The amount of energy you just calculated, should as well be discharged from the resistor R, through the charging process, while the same amount of energy, is stored in the capacitor, through this latter process. Under these circumstances, how many units of energy one should tap at the source, while charging the capacitor, to store, 1 unit of enegy on the capacitor? d) Calculate E for C=1 mikrofarad and V(0=10 volt.