Capacitor and Time Constant

We started the day with an expertiment

Understanding the Capacitor system:

When we make a circuit with battery and capacitor placed in parallel to each other, we want to see what would happen to the light bulb. When we connect the switch, nothing happens. When we take off the battery from the circuit and connect the switch again, the bulb lights up for a little while, but eventually turn off. This shows that the energy from the battery stored in a capacitor, then the energy in the capacitor is used in the circuit. We then predict the graph of brightness vs. time and we found the graph to be inverse natural log.

Understanding Time Constant:

Time constant is the amount of time for a constant to charge. Time constant is proportional to resistance and capacitor. Thus, time constant is equals to resistance multiplied by capacitor.

We then separate into 2 groups, each group set 3 batteries and 2 batteries in series respectively. Each of these set is connected to 1 capacitor. In the end, we set the 2 capacitors in parallel and predict the total voltage. We found the voltage to be 1.536 V. Using multimeter, we found the total voltage to be ..., which is closed to what we predicted.

To prove if our prediction is correct, we use logger pro. We connect the logger pro sensor to capacitor as shown below.

We found 2 graphs, the first one is when capacitor is connected to the battery, which is shown in blue, and the other one is when the battery is taken off, which is shown in red.

We want to find the unit of time constant. We know that time constant is CR, but we also know that Capacitor is charge per voltage, and voltage is consist of current multiplied by resistor. By subtituting these equations, we can find CR to be charge per current, but charge per second. Thus, we found the time constant's unit to be second.

From this relation, we can find the equation of final charge. By using current as change of Q over time, we can replace I with this and find change of Q over time to be charge/CR. When we derive the equation, we found Q to be initial Q multiplied by exponential of t/CR.

Since we know that voltage is charge divided by capacitor, we can use this equation in terms of voltage.

Next, we know that graph of voltage vs time is exponential and brightness vs time is inverse exponential. We want to predict the graph of current vs time. We know that power is current multiplied by voltage, which means when voltage increases, power lowers, and current has to lower as well. Thus, the graph of current vs time is inverse exponential.

Given the circuit in circle below, we are given the knowns as below. First, we are asked what the time constant of the circuit when the circuit is closed. Changing all to SI unit, we plug in the knowns to the formula of constant, but the resistor effected only the top resistor because the current does not go in the middle. We can also find the charge by plugging in voltage and capacitor and multiply them.

Then, we want to find the time constant when the circuit is closed. We have to first calculate the resistance affecting the circuit because both resistances are affecting since the current is going on the top circuit. Lastly, we can calculate the time it takes for the capacitor to decrease  from 4.5 to 3. By using the equation that we derived earlier, we just plug in the knowns for time constant, initial voltage, and the ending voltage, which is 3V. We found the time to be 1.09s.

From the initial charge and time constant that we found above, we can find how long it takes for the charge to go down to 1 atom. We know that 1 atom is 1.6 x 10^(-19). By using the equation that we found earlier, we found the time to be about 149s, or 2.5 min.