Lab 4Problem 5: Circuits With Two Capacitors
Problem Description:
Determining how the capacitance of a capacitor is related to the rate at which the current
decreases. The time at which the current reaches a certain fraction of its original value will be recorded
for each of three circuits.
Prediction:
The rate at which the current in each circuit will decrease will be at a faster rate
for a higher capacitance in the circuit. This means that due to the equivalent capacitances of each
circuit,
circuit XI
will decrease in current at the fastest rate, followed by
circuit IX
then
circuit X.
This
relationship is demonstrated by the equation:
Our experiment will vary the equivalent capacitance in each different circuit and will show a negative
relationship between equivalent capacitance and amount of time it takes the current to decrease to each
chosen level of current measured.
**Any necessary equations not mentioned and derivations of above equations can be found in the data section.
Procedure:
First we gained access to necessary materials, which include: a several wires, clamps, two
capacitors of the same capacitance, a 6V battery, and a digital multimeter. First ensured each capacitor
was discharged, and then we set up circuit IX. Immediately after completing the circuit we measured
the initial current. Then we recorded the time it took for the current to reach each of 4 different
fractions of the original current value. We followed this same process for three trials with each of the
three circuits. This process and the resulting measurements gave us a table of time values at which the
current value reached each of the “checkpoints”.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '11
 Marshak
 Capacitance, The Circuit

Click to edit the document details