4/30/15 - Capacitors Cont.

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We start off the day by picking up some lab equipment, mainly light bulbs, batteries, and a 1 Fared capacitor.

We run various experiments to see the behaviour of the circuit when a capacitor is hooked up, in this picture, we had hooked up the amp-meter to the circuit and observed the current over time. 

Our last measurement was to see how the potential difference was measured when we charged up two capacitors with different potential and measured them parallel.

This whiteboard shows what we had predicted the the resulting potential to be from the circuit, but we noticed that our answer of an average of the two potentials was wrong. Way wrong, but we had not figured out why the result was what it was. 

Next we measure the potential with LoggerPro while we charge the capacitor. We hook up the elements to a battery array and attach the sensors to the capacitor to measure the potential while the capacitor is being charged by the battery.

Here is the result of the data collection of the potential while charging, we fit an exponential to the curve.

Here we have the graph of the potential while the capacitor is discharging, we notice it is also an exponential graph,

The professor's graphs but with both on once data collection loop, we notice that it matches ours to the extent that it is also exponential.

What we notice about the graph is that the C constant remains the same in both expoentials. While we have represented the other constants with Voltage, we found that the constant must be of the units of 1 over time.

We then used our definition of capacitance, voltage, and resistance to set our arbitrarily constant proportional to capacitance times resistance.

After, we used the definition and canceled units until we were left with just time as the unit.

Therefore, our C constant would be one over our arbitrary constant.

Since V is equal to Q / C and V is equal to IR, we set them equal. Then we define I to be dQ / dt so then we plug in to find a differential equation for the voltage. Given we integrate after separation, we obtain the result that Q = Q_0 x exp( -t / RC ).

We blew up a capacitor by overloading the voltage, The very thin separation's resistance had been overcome and the capacitor shorted.

Here we graph the current as a function of time, and we found that the current is a downward exponential curve when it is charging, furthermore, we found an equation that relates to the graph.

Given a problem, we find the time constant for the top and middle branch. Since the middle branch has no capacitance, the time constant is zero.

Since the voltage is split in a series, we found an equation to find the voltage at any given time during the circuit. We then plugged in to find how long it would take for the potential difference to reach a certain value.

How we use the discharging equation to find how long it would take for the charge left on the capacitor to be 1 electron.

We found the answer to be just over two minutes, however to completely drain the plates of charge, it would take an infinite amount of time.