When you connect an uncharged capacitor and a resistor in series to a battery, the voltage drop is initially all across the resistor. The voltage drop across a capacitor is proportional to its charge, and it is uncharged at the beginning; whereas the voltage across the resistor is proportinal to the current and there is a current at the start. But charge starts to build up on the capacitor, so some voltage is dropped across the capacitor now. With less voltage being dropped across the resistor, the current drops off. With less current, the rate at which charge goes onto the capacitor decreases. The charge continues to build up, but the rate of that build up continues to decrease. In mathematical language, the charge as a function of time Q(t) increases but its slope decreases.

Theory says the charge obeys

Q(t) = C V_{S} (1 - e^{-t/τ})

where t is the time variable, C is the capacitance, V_{S} is the saturation voltage which in this case is equal to
the voltage across the battery, and τ=RC, a
time constant.
The corresponding voltage across the capacitor is

V(t) = V_{S} (1 - e^{-t/τ})

Now consider the case of a charged capacitor, a resistor and no battery. With no battery to "push" the charges around, the opposite charges on the capacitor plates would prefer to be together. They must pass through the resistor before they can reunite. With all those like charges on one plate, there is a strong incentive for charges to leave the plate. However, as charges leave the plate, this incentive decreases, thus the rate at which charges leave decreases as well. In mathematical language, this time the charge as a function of time Q(t) decreases and its slope decreases.

Theory says the charge obeys

Q(t) = Q_{0}
e^{-t/τ}

where t is the time, Q_{0} is whatever charge we
happen to start with, and τ=RC
is the same time constant as above. The corresponding
voltage across the capacitor is

V(t) = V_{0}
e^{-t/τ}

where V_{0} is the initial voltage.

- Connect your interface to a power amplifier and set it up as a DC voltage source. (Recall DC is at the top of the drop-down list.)
- Connect a voltage sensor (shown below) to analog channel B.

- Then in Data Studio click on the analog plug icon into the analog channel B icon and choose "voltage sensor." Drag a graph into analog channel B so we can see the data as it comes in.
- On the Signal Generator (power amplifier) set the voltage to 5.00 V.
- Make a circuit with resistor R
_{A}and a capacitor in series (have the power amplifier "off"). Your capacitor should be discharged. Diagrammatically this circuit is represented by

- Put the leads of the voltage sensor across the capacitor.
Your circuit will look something like the following

The green wire represents the RC circuit (resistor and capacitor in series), while the black and red wires represent the voltage probe. - Click
**Start**to turn on the voltage and start recording data. - You should see the voltage increase and "saturate" at 5.00 V. When it is fairly close to 5.00 V, stop recording, disconnect
the capacitor and then turn off the signal generator.
**You must disconnect first so that the capacitor will have a charge left on it!** - Table the data; copy and paste the data over into Excel.
- The charging capacitor data should be described by the
equation

V(t) = V_{S}[1 - exp( - t / τ ) ],

where V_{S}is the saturation voltage, τ is a characteristic time, and where it is assumed that t=0 is when we switched on the voltage supply. - If you started recording before switching on the signal generator, you must adjust the time. (Ask for help if this is the case with your data.)
- Plot
*Voltage vs Time* - At this stage follow these
instructions for fitting the data.

Charging R_{A}V _{S}( )τ ( )

- With the voltage sensor across the now charged capacitor, begin recording a new set of data.
- Connect the capacitor to resistor R
_{A}(no power supply). - The voltage should start close to 5.00 V and decay to zero.
Diagrammatically the circuit looks like the following

- Copy and paste the data (times and voltages) into Excel.
- For this case
you can use the
option in Add trendline. Compare the τ's.**exponential**

Discharging R_{A}V _{0}( )τ ( )

- Repeat the charging part of the above measurements with
resistor R
_{B}. - Then repeat the discharging part of the above
measurements with the combination of R
_{A}and R_{B}in series. - Do the plotting as above and compare τ's.
- Does the result agree with theory?

Charging R_{B}V _{S}( )τ ( )

Discharging R_{A}and R_{B}V _{0}( )τ ( ) Theory τ( ) Percent Error

- Get the measurements of the resistances R
_{A}and R_{B}using the multimeter as an ohm-meter. - Use your results to extract the capacitance. Note this
is the first time we have a capacitance as opposed to a ratio
of capacitances.
R _{A}( )R _{B}( )

Time constant ( ) Capacitance ( ) Charging with Resistor A Discharging with Resistor A Charging with Resistor B Discharging with Resistors A and B in series Average ( ) St. dev. ( )

Use the theory of equivalent capacitance and equivalent resistance

1 / C_{series} = 1 / C_{1} + 1 / C_{2}

C_{parallel} = C_{1} + C_{2}

R_{series} = R_{1} + R_{2}

1 / R_{parallel} = 1 / R_{1} + 1 / R_{2}

to answer the following questions. Include in your report:

- What is your best estimate of the time constant if the resistors in the part-II discharging had been in parallel?
- What would happen to the time constant (increase or decrease) if we replaced the single capacitor with the combination of it and a second capacitor in series? Why?
- What would happen to the time constant (increase or decrease) if we replaced the single capacitor with the combination of it and a second capacitor in parallel? Why?