trekman Posted August 5, 2004 Report Share Posted August 5, 2004 I must be dense, but my college book gives me this question, but as I look throughout the book, I can't find how to solve this problem."If a charge of 1 microcoulomb is placed in a 1 microfarad capacitor and then the charged capacitor is connected across a 10 ohm resistance, current will begin to flow in the resistance of?The choices it gives are10 milliamps100 milliamps1000 milliamps2 amps5 ampsA formula would be helpful. Quote Link to comment Share on other sites More sharing options...
surajbarkale Posted August 5, 2004 Report Share Posted August 5, 2004 Q = C * Vthus,V = Q / C = 1u / 1u = 1 V& According to Mr. OhmI in resistor = 1 / 10 = 100mA Quote Link to comment Share on other sites More sharing options...
Kevin Weddle Posted August 8, 2004 Report Share Posted August 8, 2004 I = C dv/dt is the equation I use most often. I also use E = 1/2 CVsquared. So I = C * 1/RC = 1/10 = 100mANow the current goes to zero, so you can increase it to 5RC and the current goes toward zero. Interesting enough is that dv/dt is a number that is associated with frequency when doing calculations. Frequency is based on the time of 1cycle where the RC is the time of a discharge which is not really a cycle. It is also interesting that the formula for capacitive reactance has dv/dt which gets cancelled out and your left with frequency which is inverse the time of the cycle. Could somebody explain why they use frequency and not dv/dt? I think it is because the instantaneous value would give it another dimension, which would be voltage. You can't easily calculate impedance when the amplitude change would would create a different dv/dt. Quote Link to comment Share on other sites More sharing options...
audioguru Posted October 2, 2004 Report Share Posted October 2, 2004 Hi Guys,In practice I just look at a graph that I have. It shows that a capacitor charges exponentially through a resistor from a constant voltage source to 63% of the supply voltage, in a time that equals RC. When fully charged (99% charged in 5RC), it discharges to 37% in the same single RC time period. The graph shows all other time constants and percentages of voltage.For the 3dB down (0.707 of V) frequency of an RC filter, I don't bother with 1 over 2 X Pie, I just simply divide 0.16 by RC. Close enough. Quote Link to comment Share on other sites More sharing options...
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