Problem: A capacitor holds charge at 9.771 `microC/Volt. It is in series with a resistance of 7.070 Ohms. The capacitor is charged to 32.57 `microC, then the source is removed, allowing the capacitor to discharge through the resistor. How much current will be flowing at the instant the capacitor begins to discharge? How long will it take the capacitor to release 1% of its charge? How much energy will be discharged during this time?
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Solution: A capacitor which holds 9.771 `microC/Volt, when holding 32.57 `microC, will be at potential ( 32.57 `microC)/( 9.771 `microC/Volt) = 3.333 Volts. At the instant the source is removed, this will be the voltage across the resistor. This voltage across a 7.070 ohm resistor will result in a current of ( 3.333 V)/( 7.070 ohm) = .4714 Amps = .4714 C/s. Now, we desire to remove 1% of the 32.57 `microC, or .3257 `microC = .0000003257 C, at .4714 C/s. This will take ( .0000003257 C) / ( .4714 C/s) = .0000006909 seconds. As this .3257 `microC = .0000003257 C flows across the 3.333 Volt = 3.333 J/C potential difference, the energy dissipated will be ( .0000003257 C)( 3.333 J/C) = .000001085 Joules.
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.
