Problem: A capacitor holds charge at 6.187 `microC/Volt. It is in series with a resistance of 1.473 Ohms and a 9.918 volt source. When the capacitor holds a charge of 53.62 `microC, approximately how long will it take it to increase its charge by 1%?
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Solution: A capacitor which holds 6.187 `microC/Volt, when holding 53.62 `microC, will be at a potential of ( 53.62 `microC)/( 6.187 `microC/Volt) = 8.666 Volts. This voltage will oppose that of the source, so that the total voltage across the resistor will be the source's 9.918 volts, less the 8.666 Volts across the capacitor, or 1.252 Volts. This voltage across a 1.473 ohm resistor will result in a current of ( 1.252 V)/( 1.473 ohm) = .8499 Amps = .8499 C/s. Now, we desire to add 1% of the 53.62 `microC, or .5362 `microC = .0000005362 C, at .8499 C/s. This will take ( .0000005362 C) / ( .8499 C/s) = .0000006308 seconds. Note that as charge builds, the voltage across the resistor decreases and current decreases, so that it takes longer and longer to increase charge by a given amount; however, it always takes the same time to build the charge on a given capacitor-resistor combination by the same FACTOR. Try it. Rework the problem, changing only the amount of charge on the capacitor (just don't give it so much charge that its voltage exceeds that of the source; since its voltage comes from the source, this can't happen).
Generalized Response:
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Figure description: The figure below charts the relationships among voltage, resistance, current and power. The 'blue' triangle shows how voltage, current and resistance are related. (Greater voltage or less resistance implies greater current, reflecting I = V / R; resistance is the ratio of voltage to current, reflecting R = V / I; a current I through a resistance R requires a greater voltage drop pfor a greater current and for a greater resistance, reflecting V = I * R).
The 'green' triangle shows how power is the product of current and voltage (current is measured in C / s, voltage in J / C, so the product of current and voltage is the number of J / s, or watts, of power).
The relationship P = I * V can be combined with either I = V / R or with V = I * R to yield either P = I * (I * R) = I^2 * R or P = (V / R ) * V = V^2 / R.
