Question

A charged capacitor is connected to a resistor and switch, all in series. The circuit has a time constant T = 1.80 s. Soon after the switch is closed, the charge on the capacitor is 65.0% of its initial charge qu = 100°C. a) Find the time interval required for the capacitor to reach this charge. b) If R = 210 22, what is the value of C? c) What is the current when the charge on the capacitor is 65.0% of its initial charge? d) What is the power dissipated in the resistor at that instant? e) Plot charge on the capacitor and current in the circuit with respect to time. It doesn't have to be to scale, just a sketch.

Answer 1

at 9 - 10onc T = 1.80 sec charge in capacitor any -t/e given by obitary time t is q=% (1-ethe q=65% of 90= 0.65 q. 0.65% = %(1-e tri gost) tilos e 1-0.65 0.35 -Higos) ence en (0-35) -t 1.80s --1,049 + t=(-1.049)(11803)= 1.89 sec t-1.89 sec T=Rc =) = T R 1.80s 2102 8.57X16°F Ca 8.57X103F while discharging i= -dq dt - the i= %. て e at t 1.89 sec iz B 1.89%elos looMC 1.80 sec e i. 100x1560 XO-35 1.80 sec + i= 1.944 x10 SA Power dissipated in resistor P = (²R P. - (1.944416-5a)? (2002) P- 7.94x108 W 90 it Scanned with CamScanner