Basic Electrical Engineering Lab, Experiment 4: Time constant of an RC circuit
Course: EE 151 Basic Electrical Engineering Lab · Experiment: 4 · Author: Kavya Reddy · Date: 16 Oct 2026
| Group | Members | Bench | Instructor |
|---|---|---|---|
| B3 | Kavya Reddy (23EE041), Rohan Das (23EE057), Sana Qureshi (23EE062) | 7 | Dr. P. Natarajan |
Aim
To measure the time constant of a series RC circuit from its charging curve and compare it with the value \tau = RC from the component ratings.
Apparatus
| Item | Specification | Quantity |
|---|---|---|
| Regulated DC supply | 0 to 30 V, set to 5.00 V | 1 |
| Resistor | 10 kΩ, 1% carbon film (measured 9.96 kΩ) | 1 |
| Electrolytic capacitor | 470 µF, 25 V, tolerance 20% | 1 |
| Digital multimeter | 20 V range, least count 0.01 V | 1 |
| Stopwatch | Least count 0.01 s | 1 |
| SPDT switch, breadboard, leads | 1 set |
Theory
When a DC voltage V_0 is applied to an uncharged capacitor through a resistor, the capacitor voltage rises as
v_C(t) = V_0\left(1 - e^{-t/\tau}\right), \qquad \tau = RC
At t = \tau the capacitor reaches 1 - e^{-1} = 63.2\% of V_0. Rearranging and taking logs gives a straight line through the origin:
-\ln\left(1 - \frac{v_C}{V_0}\right) = \frac{t}{\tau}
so the slope of -\ln(1 - v_C/V_0) against t is 1/\tau. With the rated values, \tau = 10 \times 10^3 \times 470 \times 10^{-6} = 4.70 s.
Setup
flowchart LR
S["5 V DC supply"] --> SW["SPDT switch"]
SW --> R["R = 10 kΩ"]
R --> C["C = 470 µF"]
C --> G["Ground"]
C -.- M["Multimeter across C"]
Procedure
- Discharge the capacitor fully by shorting it through the switch's second position for 30 seconds.
- Connect the circuit as shown, with the multimeter across the capacitor on the 20 V range.
- Set the supply to 5.00 V and confirm it on the multimeter before connecting.
- Throw the switch to the supply and start the stopwatch at the same moment.
- Record the capacitor voltage at 2, 4, 6, 10 and 15 seconds, and the time at which it reads 3.16 V (63.2% of 5.00 V).
- Repeat steps 1 to 5 twice more and average the readings.
Precautions
- Observe the capacitor's polarity. A reversed electrolytic capacitor can fail violently.
- Discharge the capacitor fully before every run, or the curve starts above zero.
- Keep the multimeter range fixed during a run so its input resistance stays the same.
Observations
Supply voltage V_0 = 5.00 V. Each voltage is the mean of three runs.
| Time (s) | v_C (V) |
1 - v_C/V_0 |
-\ln(1 - v_C/V_0) |
|---|---|---|---|
| 2 | 1.67 | 0.666 | 0.406 |
| 4 | 2.80 | 0.440 | 0.821 |
| 6 | 3.54 | 0.292 | 1.231 |
| 10 | 4.34 | 0.132 | 2.025 |
| 15 | 4.77 | 0.046 | 3.079 |
Time to reach 3.16 V: 4.88 s, 4.93 s and 4.90 s, mean 4.90 s.
View data
| seconds | volts |
|---|---|
| 0 | 0 |
| 2 | 1.67 |
| 4 | 2.8 |
| 6 | 3.54 |
| 10 | 4.34 |
| 15 | 4.77 |
Calculations
Sample calculation for t = 10 s:
1 - \frac{4.34}{5.00} = 0.132, \qquad -\ln(0.132) = 2.025, \qquad \frac{t}{2.025} = \frac{10}{2.025} = 4.94 \text{ s}
A least-squares line through the origin over all five points gives the slope
m = \frac{\sum t_i y_i}{\sum t_i^2} = \frac{77.92}{381} = 0.2045 \text{ s}^{-1} \quad\Rightarrow\quad \tau = \frac{1}{m} = 4.89 \text{ s}
The direct 63.2% method gave 4.90 s, which agrees with the slope method to within 0.01 s.
Result
Measured time constant \tau = 4.89 \pm 0.10 s. Expected from the ratings: 4.70 s.
Error analysis
\text{Percentage error} = \frac{|4.89 - 4.70|}{4.70} \times 100 = 4.0\%
- Capacitor tolerance (dominant). The capacitor is rated 470 µF with a 20% tolerance. With the measured resistance of 9.96 kΩ, our
\tauimpliesC = 4.89 / 9960 = 491µF, 4.5% above nominal and well inside the tolerance. - Reaction time. Starting and stopping the stopwatch by hand adds about 0.2 s per reading. Averaging three runs and fitting five points reduces its effect on
\tauto about 0.1 s. - Meter loading. The multimeter's 10 MΩ input resistance is 1000 times
R, so it changes\tauby about 0.1%, which is negligible.
Conclusion
The capacitor voltage followed the expected exponential curve, and the log plot was a straight line through the origin. The measured time constant of 4.89 s is 4.0% above the rated value, a difference explained by the capacitor's tolerance. The aim was met.
Viva questions
Why does the capacitor reach only 63.2% of the supply after one time constant?
Because 1 - e^{-1} = 0.632. The rate of charging is proportional to the remaining gap, so the voltage approaches V_0 exponentially and never quite reaches it.
How long until the capacitor is practically fully charged?
About 5\tau, roughly 24.5 s here, when it is within 0.7% of V_0.
What would doubling the resistance do?
It would double \tau, so the curve keeps the same shape but takes twice as long. The final voltage stays the same.