What it shows
A capacitor charges quickly at first and then more slowly, following an exponential curve. The time constant RC is the time to reach about 63 percent of the supply voltage.
The science
The charging current is (V minus Vc) / R, so as the capacitor voltage Vc rises, the current falls. The result is Vc = V x (1 minus e to the power minus t / RC). For R = 100 kΩ and C = 1000 µF, RC = 100 s, so after 100 s the capacitor reaches about 63 percent of 9 V, which is 5.7 V, and after 200 s about 86 percent. Real capacitors have a leakage current and a tolerance of 20 percent so the curve will be slightly lower.
Materials
Cost: under Rs 500. Many items come from home or from waste; spare parts such as LEDs, motors and wire are in the store.
- A 1000 µF 16 V electrolytic capacitor
- A 100 kΩ resistor
- A 9 V battery with snap connector
- A multimeter on DC volts
- A stopwatch or phone
- A 1 kΩ resistor for discharging the capacitor safely
- A breadboard, wires, a push button and graph paper
Safety
For this project: Battery power only. Fit the electrolytic capacitor with its negative leg to ground. Discharge it with a short wire through a 1 kΩ resistor, not with a bare short circuit.
Never use mains electricity: power models with cells, batteries or a low-voltage adapter. Sharp and hot tools, flames and lab chemicals need an adult or your teacher beside you. Keep magnets, button cells and small parts away from young children, and wash your hands after making.
Step by step
8 steps. Do them in order, and test each part before you go on to the next. Write what you notice in a notebook as you go.
Fit the capacitor with its negative leg to the ground rail and check with the meter that it reads 0 V; discharge it through 1 kΩ if not.
Connect the 100 kΩ resistor from the push button to the positive leg of the capacitor, and the other side of the button to the 9 V rail.
Connect the multimeter in volts mode across the capacitor.
Press and hold the button and start the stopwatch at the same moment.
Read the voltage every 20 seconds for 5 minutes and write the values in a table.
Calculate the theoretical voltage at each time using the formula for RC = 100 s.
Plot both on one graph with time on the horizontal axis and voltage on the vertical.
Mark the time the measured voltage reaches 5.7 V and compare it with RC.
What you should see
The graph rises steeply and then flattens; the voltage reaches about 63 percent of the supply near 100 seconds, with small differences from theory.
Viva questions
A teacher or a judge will ask how it works. Try to answer each question before you open it, then compare.
01What is the time constant?
RC, the time for the voltage to reach about 63 percent of its final value.
02Why does charging slow down?
The voltage across the resistor, and so the current, falls as the capacitor charges.
03What happens if the resistor is doubled?
The time constant doubles, so charging is twice as slow.
04Why does the measured curve differ from theory?
Tolerance of the capacitor and resistor, leakage and the meter loading the circuit.
Take it further
Discharge the capacitor through the same resistor and plot the falling curve, then compare half-life with 0.69 RC.
Change one thing, measure what happens, and add a table or chart of your readings: that is what turns a model into your own investigation.
Free ebooks for this project
Hardware & repair
Using a Digital Multimeter
Measure voltage, current, resistance and continuity safely
Robotics
Electronic Components Handbook
Resistors, capacitors, LEDs, diodes, transistors and more, with how to use each
Find out more about electronics and robotics
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