# Capacitors and derivations, RC circuits

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I am confused on specific calculations only. the derivations.

scan003.jpg part 1 a). the derivation

scan003.jpg part 1 c). the derivation

scan0004.jpg a). the derivation

Please include as many steps as possible so I can understand the process.

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The solution proves excellent explanation of the process and formulas used to complete the problems.

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1. (a) This is an important basic thing in engineering and science. What you have here is the simplest differential equation - the equation you get when some quantity is increasing or decreasing over time, and the rate of increase or decrease is proportional to the amount of the quantity at the instant in question. This applies to population, where the number of babies born is proportional to the number of people alive at the moment, etc.

In this case, when the switch is closed, the capacitor charges up to the full voltage on the battery Vs. When you open the switch, the capacitor discharges through the resistor. The current through the resistor is I = V/R, and V, the voltage, is the voltage on the capacitor. Now for a capacitor we have I = C dV/dt, which means that the current through a capacitor is equal to the capacitance (in Farads) times the rate of change of the voltage (in Volts/second). So with the two equations (I = V/R, and I = C dV/dt), and that fact that once the switch is opened the current through the capacitor must equal the current through the resistor (there's nowhere else for it to flow), but it's in the opposite direction with respect to the voltage on the capacitor and resistor, so there's a minus sign. And we get the standard first-order differential equation:

We solve this for VC by multiplying both sides by dt and integrating both sides:

Then we move the R and the VC over and get:

Then we integrate both sides, and we know that the integral of dx/x is ln(x):

where B is a constant. Then we exponentiate both sides:

Now we solve for VC:

here A is a ...

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