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. Understand and analyze the mathematics: Both differential equations are \"separable,\" first-order differential equations. If you remember your calculus, you can solve these if you know a few small tricks. Fill in the steps that are missing and annotate your work to make sure you understand this as best you can. w 6. dQ [V, O ' - | e Write your equation for the charging circuit as: "' R R e : 9] T e Put all the charge stuff on one side and the time stuff on the other: R RC I-- ('] du dt == = =- Define R RC and you will see that your equationis RC | and you can integrate both sides. Integrating the left side gives a natural log. We need a boundary condition (initial value) to determine the constant = _ +In RC In' L - J!\\ R(A V o ' of integration. Assume no charge at time equals zero. Confirm that you will have R The trick here is to take the two natural logs and put them on the same side of the equation, and use the fact that the difference in natural logs is the natural log of the ratio. Then you exponentiate both sides. This is the harder part that you might not be as familiar with, so let's walk through it. O j O Vv = R / In| =- |-In|=(=In K _KC = R RC R f__ R( R using some properties of natural logarithms I O | R RC | Te I,' | % by taking exponential of both sides, then I o Vv - I O Vo -l . - . = B R . S| =Rt Qo=VC|l-e* R RC R R RC R with a little algebra implies ; then if we use g=re for the potential difference on the capacitor, canceling the \"C\