Set 53 Problem number 6

Problem

A capacitor holds charge 9 `microC/Volt. It is in series with a resistance of 29 Ohms.

The capacitor is charged to 54 `microC, then the source is removed and the circuit again closed, 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?

Solution

A capacitor which holds 9 `microC/Volt, when holding 54 `microC, will be at potential ( 54 `microC)/( 9 `microC/Volt) = 6 Volts.

• At the instant the source is removed, this will be the voltage across the resistor.
• This voltage across a 29 ohm resistor will result in a current of ( 6 V)/( 29 ohm) = .2068 Amps = .2068 C/s.

Now, we desire to remove 1% of the 54 `microC, or .54 `microC = 5.4E-07 C, at .2068 C/s.

• This will take ( 5.4E-07 C) / ( .2068 C/s) = `k seconds.
• As this .54 `microC = 5.4E-07 C flows across the 6 Volt = 6 J/C potential difference, the energy dissipated will be ( 5.4E-07 C)( 6 J/C) = 3.24E-06 Joules.

Generalized Solution

A capacitor with capacitance C, holding charge Q, will be at voltage Q / C.  If it discharges through a resistance R, then current will flow at the rate I = Vcap / R = Q / (R C).

• To lost .01 of the charge, or .01 Q, at the rate I = dQ / dt = Q / (R C) will require approximate time `dt = .01 Q / I = .01 Q / (Q / RC) = .01 RC.
• The actual time will be slightly longer, since the current will be decreasing as the capacitor loses charge and hence experiences a decrease in voltage.