Problem: A capacitor holds charge at 8.494 `microC/Volt. It is in series with a resistance of .9333 Ohms and a 97 volt source. When the current has built a charge of 3.261 `microC on the capacitor, how much current will be flowing?
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Solution: A capacitor which holds 8.494 `microC/Volt, when holding 3.261 `microC, will be at a potential difference of ( 3.261 `microC)/( 8.494 `microC/Volt) = .3839 Volts. Since this voltage results from charge built by the current on the capacitor, it will oppose that of the source, so that the total voltage across the resistor will be the 97 volts of the source, less the .3839 Volts across the capacitor, or 96.61 Volts. This voltage across a .9333 ohm resistor will result in a current of ( 96.61 V)/( .9333 ohm) = 103.5 Amps.
Generalized Response:
.
.
.
.
.
.
.
.
.
.
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.
