If a circuit has 3.9kΩ resistor & 5µF capacitor, find current flow at 0.005 sec. if the max. current is 1.5mA

Question

The equation is I = Imax e-t/RC where (I) is the current at time (t), (Imax) is the maximum current, (R) is the resistor and (C) is the capacitor. I don't know what the letter (e) stands for but it is raised to the negative t/RC. I don't know how to solve for this kind of equation, the book didn't give any example on how to do it. Can anyone help me with this? Thanks a lot!

Answer 1

Obviously you have some catching up to do. e is the base of the natural logarithms and is involved in all aspects of exponential growth and decay. I suggest you google "natural logarithm", perhaps starting with the ref., to learn about this. The actual value of e is an irrational number, 2.7182818... With calculators widely available, it is hardly ever necessary to know the actual value of e, as there will be a button "e^x" or "exp" to do the exponentiation of e^x.
e^(+t/K) is the factor of exponential growth and e^(-t/K) is the factor of exponential decay. This problem is about exponential decay, where K is the time constant of the RC circuit, which is simply R*C = 0.0195 sec, and the decay is the decrease of capacitor charging current toward zero as the voltage on the capacitor approaches the source voltage. With t = 0.005 sec, the equation is I = 0.0015*e^-(0.005/0.0195) = 0.001161 A.

Answer 2

imagine of a river operating down a hill. Voltage is the right of the hill. cutting-edge is the speed of pass of water (e.g. litres in conserving with 2d). Resistance is what's preventing the pass of water. imagine of it as a constriction contained in the river. If there are 3 constrictions one after yet another, then the pass of water slows down 3 situations. yet when the three constrictions are in parallel, like the river breaking as a lot as 3 steams, then reforming right into a river back, then the pass is slowed yet a lot less. the answer to a) is the resistors are all in parallel. i do no longer see the diagram so i will go away it as that.