Here is a schematic sketch of the RC circuits we will be considering in this lab. The switch allows us to charge the capacitor (when the switch is closed to the a-b position) or discharge the capacitor (when the switch is closed to the b-c position). Note that when the switch is in the b-c position, the lack of a connection from the top of the power supply means that the left side of the circuit is disconnected. T xe e % \\ R C v Understand the physics: We are going to use Kirchoff's loop rule of the potential gains and losses to find an equation that we can solve to determine the charge on the capacitor. Charging For the charging loop (setting a-b), assume the current goes clockwise. Note that the +Q goes on the capacitor plate connected to the positive terminal of the power supply, so in Kirchoff's loop rule, we will subtract the potential on the capacitor. Write out the loop rule, using V. as the potential drop on the capacitor. Start at \"a\" in the diagram and go clockwise. Then replace V. using Q=VC and the current over the resistor with | = dQ/dt. Your result is a differential equation for the charge on the capacitor as a function of time. Discharging For the discharging loop (setting b-c), assume the current goes clockwise. Note that since the +Q is on the \"top\" end of the capacitor plate - the end closer to the switch - a clockwise loop will see the capacitor as adding potential in Kirchoff's loop rule. Write out the loop rule, using V. as the potential drop on the capacitor. Start at \"c\" in the diagram and go clockwise. Then replace V. using Q=VC. There is one wrinkle here for the resistor. You want the current to be positive, but the time rate of change of the charge on the capacitor is negative. (The cap is losing charge.) So when you replace the current in V= IR over the resistor, you need to use | = - dQ/dt. Your result is a slightly different differential equation for the charge on the capacitor as a function of time