Set 53 Problem number 5

Problem

A capacitor holds charge 3 `microC/Volt. It is in series with a resistance of 5440000 Ohms and a 31 volt source. When the capacitor holds a charge of 33 `microC, approximately how long will it take it to increase its charge by 1%?

Solution

A capacitor which holds 3 `microC/Volt, when holding 33 `microC, will be at a potential of ( 33 `microC)/( 3 `microC/Volt) = 11 Volts.

• This voltage will oppose that of the source, so that the total voltage across the resistor will be the source's 31 volts, less the 11 Volts across the capacitor, or 20 Volts.
• This voltage across a 5440000 ohm resistor will result in a current of ( 20 V)/( 5440000 ohm) = 3.676E-06 Amps = 3.676E-06 C/s.

Now, we desire to add 1% of the 33 `microC, or .33 `microC = 3.3E-07 C, at approximately 3.676E-06 C/s.

• This will take ( 3.3E-07 C) / ( 3.676E-06 C/s) = .08977 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).  You will obtain the same result for your estimate of the time required to increase the charge by 1%.

Generalized Solution

A capacitor is characterized by its capacitance C, which is its charge / voltage ratio. 

• If this capacitor holds charge Q, then its voltage Vcap is such that C = Q / Vcap, so that Vcap = Q / C.

When the charge is built by the current from the source, its voltage will resist that of the source, since the positive terminal of the source tends to move positive charges to the plate of the capacitor to which it is most directly connected and the negative terminal to the plate connected to it.

• Thus if Vs is the voltage of the source, the voltage across the resistor will be Vr = Vs - Vcap.
• It follows that the current through the resistor is I = (Vs - Vcap) / R = (Vs - Q / C) / R.

Note that as charge builds on the capacitor, the voltage across the resistor decreases and so therefore does the current.  This causes charge to build on the capacitor more and more slowly.

To increase the charge by 1%, we must add charge .01 Q.

• Current flows at rate I = (Vs - Q / C) / R.
• Current is the rate I = dQ / dt at which charge flows through the circuit.  Since the charge ends up on the capacitor, this is the rate at which charge is added to the capacitor.
• To add charge .01 Q at this rate will require time `dt = .1 Q / I = .1 Q / [ (Vs - Q / C) / R ] = .01Q * R / (Vs - Q / C).