# Charging and Discharging a Single Capacitor

*RC Circuit: Charging and Discharging a Single Capacitor*

We wish to make plots of the time-dependent behavior of capacitor voltage and current of the circuit. Take ten *t*, *V _{C}*, and

*i(t)*data points from the simulation. For the voltage plots, we wish to investigate what occurs when the capacitor is charging/discharging after closing a switch. Be sure to follow the conventions that are defined in section 2 of the lab manual – this should ensure what quantities go on each respective axis.

**Voltage plots:**

1. Using data from the simulation for the capacitor charging, plot the following:

2. Using data from the simulation for the capacitor discharging, plot the following:

**Current plots:**

3. Using data from the simulation for the capacitor charging, plot the following:

Show that the calculation of ?? agrees with the given values for resistance and capacitance in the circuit by considering the following:

4. **Fitting the semi-log plots of v(t) and i(t) on the spreadsheet.** After fitting the data with a trendline, we can extract the slope and comment on how this will be used to determine the time constant, ??.

I [mA] |
VC (V) |

45.441 |
0.066 |

30.561 |
1.682 |

20.021 |
2.827 |

13.228 |
3.558 |

8.819 |
4.043 |

5.775 |
4.373 |

3.884 |
4.578 |

2.544 |
4.724 |

The semi-log slope of Vc and Ic has to be 1 but the experimental slope is almost 1, so we have to find time constant from Vc and Ic separately after that we have to average these two values.

From the equation 1/time_constant = Slope of this plot - charging

From the equation 1/time_constant = Slope of this plot – Charging

**Measurement of t _{1/2} and calculation of **

**??.**In a few sentences, describe how your experimental (simulation) value of ?? corresponds to the actual value.

I use the slope of the voltage and current value for finding time constant.

Time Constant value is between 10^{-3}/0.0413 and 10^{-3}/0.0413

10^{-3}/0.0413 = 0.02421

10^{-3}/0.0413 = 0.02421

**Average value of ****??**** is 0.02421**

**Average value of t _{1/2} is 0.016781**