A capacitor charging and discharging experiment observes the charging process in which charge is stored in a capacitor through a resistor and the discharging process in which the stored charge is released through a resistor, examining how voltage, current, and charge change with time.
The voltage across a capacitor does not change instantaneously immediately after connection to a power supply; instead, it gradually increases with time. Likewise, when a charged capacitor is discharged, its voltage gradually decreases with time.
These changes are described by exponential functions, and their rate is determined by the time constant τ = RC, the product of resistance R and capacitance C.
This article assumes a charging and discharging experiment using a DC power supply, resistor, capacitor, switch, and voltmeter or oscilloscope, and presents an introduction, precautions, result items, sample experimental values, calculations, sources of error, example result writing, and discussion examples.
Note:
This article is a study reference for discussing results obtained in physics experiments at schools, universities, and similar institutions. The experimental conditions and measured values shown here are illustrative examples and are not actual measurement results.
Electrolytic capacitors are polarized. Connecting one in reverse or exceeding its rated voltage can cause heating, leakage, or damage. Turn off the power before changing the wiring and confirm that no charge remains in the capacitor. Do not short-circuit a capacitor directly with a wire for rapid discharge; discharge it through an appropriate resistor. Follow your institution’s laboratory manual and the instructions of your instructor or TA.
- What Is Capacitor Charging and Discharging?
- Main Items to Record in the Results
- Sample Experimental Conditions
- Sample Experimental Values
- Calculation Method
- Calculating the Theoretical Time Constant
- Calculating the Maximum Charge
- Voltage After 1 Second of Charging
- Charge After 1 Second of Charging
- Current After 1 Second of Charging
- Voltage After 1 Second of Discharging
- Obtaining the Time Constant from 63.2%
- Obtaining the Time Constant from 36.8%
- Linearizing the Discharge Curve
- Calculating Capacitance from the Experimental Time Constant
- Energy Stored in the Capacitor
- Relative Error
- Main Sources of Error
- Why the Current Is Maximum Immediately After Charging Begins
- Why the Capacitor Voltage Does Not Change Instantaneously
- Why a Larger Resistance Slows Charging and Discharging
- Why a Larger Capacitance Slows Charging and Discharging
- Why the Discharge Current Flows Opposite to the Charging Current
- Why the Process Is Considered Complete After About Five Time Constants
- Example of How to Write the Results
- Points to Connect to the Discussion
- Discussion Example
- Summary
What Is Capacitor Charging and Discharging?
A capacitor is a component that stores electric charge between two electrodes. The charge Q, capacitance C, and voltage V across a capacitor are related as follows.
Q = CV
A larger capacitance allows more charge to be stored at the same voltage.
Capacitor Voltage During Charging
When an initially uncharged capacitor is connected through a resistor to a DC source of voltage V0, the capacitor voltage is:
VC(t) = V0{1 − exp(−t/RC)}
As time passes, the capacitor voltage approaches the supply voltage.
Charge During Charging
Q(t) = CV0{1 − exp(−t/RC)}
After sufficient time, the maximum charge is:
Qmax = CV0
Current During Charging
I(t) = (V0/R)exp(−t/RC)
The current is maximum immediately after charging begins and then decreases with time.
Capacitor Voltage During Discharging
When a capacitor charged to an initial voltage V0 is connected to a resistor and discharged, its voltage is:
VC(t) = V0exp(−t/RC)
Charge During Discharging
Q(t) = Q0exp(−t/RC)
Current During Discharging
If the charging-current direction is defined as positive, the discharge current flows in the opposite direction and is written with a negative sign.
I(t) = −(V0/R)exp(−t/RC)
Time Constant
τ = RC
One time constant after charging begins, the capacitor voltage reaches about 63.2% of its final value.
VC(τ) ≈ 0.632V0
One time constant after discharging begins, the capacitor voltage falls to about 36.8% of its initial value.
VC(τ) ≈ 0.368V0
After about five time constants, charging reaches about 99.3% of the final value and discharging falls to about 0.67% of the initial value, so the process can be regarded as practically complete.
Main Items to Record in the Results
- Nominal capacitance of the capacitor
- Capacitance tolerance and rated voltage
- Capacitor polarity
- Nominal and measured resistance
- Supply voltage
- Initial voltage before charging
- Initial voltage before discharging
- Elapsed time and capacitor voltage during charging
- Elapsed time and capacitor voltage during discharging
- Charge during charging
- Current during charging and discharging
- Maximum charge
- Theoretical time constant
- Experimental time constant obtained from 63.2% and 36.8%
- Time constant obtained from the slope of a semilogarithmic graph
- Capacitance obtained from the experimental time constant
- Relative error between theoretical and experimental values
- Energy stored in the capacitor
- Oscilloscope time scale, voltage scale, and probe attenuation
Sample Experimental Conditions
| Item | Sample Condition |
|---|---|
| Resistor R | 10.0 kΩ |
| Capacitor C | 100 μF |
| Supply voltage V0 | 5.00 V |
| Theoretical time constant τ | 1.00 s |
| Maximum charge Qmax | 500 μC |
| Theoretical current at start of charging | 0.500 mA |
| Measurement method | Digital voltmeter or oscilloscope |
| Measurement time | 0–5.0 s |
| Measurement interval | 0.5 s |
| Room temperature | 23 ℃ |
Sample Experimental Values
Voltage, Charge, and Current During Charging
| Time t (s) | Measured Voltage (V) | Charge Q (μC) | Current I (mA) |
|---|---|---|---|
| 0.0 | 0.02 | 2 | 0.498 |
| 0.5 | 1.95 | 195 | 0.305 |
| 1.0 | 3.14 | 314 | 0.186 |
| 1.5 | 3.86 | 386 | 0.114 |
| 2.0 | 4.30 | 430 | 0.070 |
| 2.5 | 4.56 | 456 | 0.044 |
| 3.0 | 4.73 | 473 | 0.027 |
| 4.0 | 4.89 | 489 | 0.011 |
| 5.0 | 4.95 | 495 | 0.005 |
Voltage, Charge, and Current During Discharging
| Time t (s) | Measured Voltage (V) | Charge Q (μC) | Current Magnitude (mA) |
|---|---|---|---|
| 0.0 | 4.98 | 498 | 0.498 |
| 0.5 | 3.05 | 305 | 0.305 |
| 1.0 | 1.86 | 186 | 0.186 |
| 1.5 | 1.14 | 114 | 0.114 |
| 2.0 | 0.70 | 70 | 0.070 |
| 2.5 | 0.43 | 43 | 0.043 |
| 3.0 | 0.27 | 27 | 0.027 |
| 4.0 | 0.10 | 10 | 0.010 |
| 5.0 | 0.04 | 4 | 0.004 |
Charging Time with Different Resistance Values
| Resistance R | Capacitance C | Theoretical Time Constant RC | Experimental Time Constant | Approx. 5τ |
|---|---|---|---|---|
| 5.0 kΩ | 100 μF | 0.50 s | 0.49 s | 2.5 s |
| 10.0 kΩ | 100 μF | 1.00 s | 0.99 s | 5.0 s |
| 20.0 kΩ | 100 μF | 2.00 s | 1.97 s | 10.0 s |
Charging Time with Different Capacitances
| Resistance R | Capacitance C | Theoretical Time Constant RC | Experimental Time Constant | Maximum Charge |
|---|---|---|---|---|
| 10.0 kΩ | 47 μF | 0.47 s | 0.46 s | 235 μC |
| 10.0 kΩ | 100 μF | 1.00 s | 0.99 s | 500 μC |
| 10.0 kΩ | 220 μF | 2.20 s | 2.16 s | 1100 μC |
Calculation Method
Calculating the Theoretical Time Constant
R = 10.0 kΩ = 1.00 × 104 Ω
C = 100 μF = 1.00 × 10−4 F
τ = RC
= (1.00 × 104)(1.00 × 10−4)
= 1.00 s
Calculating the Maximum Charge
Qmax = CV0
= 100 × 10−6 × 5.00
= 5.00 × 10−4 C
= 500 μC
Voltage After 1 Second of Charging
VC(t) = V0{1 − exp(−t/RC)}
= 5.00{1 − exp(−1.00/1.00)}
= 3.16 V
Charge After 1 Second of Charging
Q = CV
= 100 × 10−6 × 3.16
= 3.16 × 10−4 C
= 316 μC
Current After 1 Second of Charging
I(t) = (V0/R)exp(−t/RC)
= (5.00/10000)exp(−1)
= 1.84 × 10−4 A
= 0.184 mA
Voltage After 1 Second of Discharging
VC(t) = V0exp(−t/RC)
= 5.00exp(−1.00/1.00)
= 1.84 V
Obtaining the Time Constant from 63.2%
For a supply voltage of 5.00 V, the voltage corresponding to 63.2% during charging is:
0.632 × 5.00 = 3.16 V
Read the time at which the charging curve reaches 3.16 V. If it is 0.99 s in this example, the experimental time constant is 0.99 s.
Obtaining the Time Constant from 36.8%
If the measured voltage at the start of discharging is 4.98 V, 36.8% of it is:
0.368 × 4.98 = 1.83 V
Read the time at which the discharge curve falls to 1.83 V and use it as the time constant.
Linearizing the Discharge Curve
VC = V0exp(−t/RC)
lnVC = lnV0 − t/RC
On a graph with time t on the horizontal axis and lnVC on the vertical axis, the slope is −1/RC.
Slope a = −1/τ
τ = −1/a
If the slope of the best-fit line is −1.01 s−1:
τ = −1/(−1.01)
= 0.990 s
Calculating Capacitance from the Experimental Time Constant
For an experimental time constant of 0.990 s and measured resistance of 10.1 kΩ:
C = τ/R
= 0.990/(10.1 × 103)
= 9.80 × 10−5 F
= 98.0 μF
Energy Stored in the Capacitor
U = (1/2)CV2
= (1/2) × 100 × 10−6 × 5.002
= 1.25 × 10−3 J
Relative Error
For a theoretical time constant of 1.00 s and experimental time constant of 0.99 s:
Relative error (%) = |Experimental value − Theoretical value| / Theoretical value × 100
= |0.99 − 1.00| / 1.00 × 100
= 1.0%
Main Sources of Error
| Source of Error | Effect on Measurement | Likely Trend in Results | Improvement |
|---|---|---|---|
| Capacitance tolerance | Actual capacitance differs from nominal value | Time constant and maximum charge deviate from theory | Determine actual capacitance from experimental results |
| Resistor tolerance | Actual resistance differs from nominal value | Charging and discharging rates change | Use measured resistance in calculations |
| Capacitor leakage current | Stored charge is lost internally | Final charging voltage is lower and discharge is faster | Use a capacitor with low leakage |
| Equivalent series resistance | Adds an internal resistance component | Initial current and short-time behavior deviate from theory | Use a component with low ESR |
| Dielectric absorption | Voltage recovers slightly after discharge | Residual voltage appears after apparent full discharge | Discharge through a resistor for sufficient time |
| Power-supply internal resistance | Increases effective resistance and lowers terminal voltage | Charging time constant increases and final voltage decreases | Measure terminal voltage under load |
| Voltmeter input resistance | Adds a parallel discharge path | Discharge time constant becomes smaller | Use a high-input-resistance instrument |
| Oscilloscope input capacitance | Increases effective capacitance | Time constant increases in low-capacitance circuits | Use a 10× probe |
| Switch contact resistance | Makes circuit resistance unstable | Time constant varies between measurements | Clean contacts and ensure secure connections |
| Manual-operation reaction time | Offsets the charging/discharging start from recording start | Time constant may be over- or underestimated | Use automatic recording or triggering |
| Residual initial charge | Initial charging voltage is not zero | Charging curve deviates from theory | Discharge sufficiently before measurement |
| Insufficient charging | Voltage before discharge has not reached supply voltage | Error in assumed initial value | Charge for at least 5τ |
| Supply-voltage fluctuation | Changes final voltage and maximum charge | Repeated curves do not coincide | Record supply voltage for each measurement |
| Coarse measurement interval | Fails to capture changes near the time constant adequately | Reading error in the 63.2% and 36.8% times | Measure more finely around 0–2τ |
| Noise at low voltage | Relative error increases late in discharge | End of semilog plot departs from a straight line | Exclude points near the noise floor from fitting |
Why the Current Is Maximum Immediately After Charging Begins
Before charging, the capacitor voltage is nearly zero, so immediately after connection almost the entire supply voltage appears across the resistor. Therefore, the initial current is maximum at V0/R.
As charging proceeds, the capacitor voltage rises and the voltage across the resistor decreases, so the current decreases.
Why the Capacitor Voltage Does Not Change Instantaneously
The charge on a capacitor is given by Q = CV. Changing the voltage instantaneously would require transferring a very large amount of charge in an infinitesimal time.
Because actual circuits have resistance, current is finite and capacitor voltage changes continuously.
Why a Larger Resistance Slows Charging and Discharging
With a larger resistance, less current flows at the same voltage. Because less charge is transferred per unit time, charging and discharging require more time.
Why a Larger Capacitance Slows Charging and Discharging
A capacitor with larger capacitance requires more charge to reach the same voltage. Therefore, with the same resistor, charging and discharging take longer.
Why the Discharge Current Flows Opposite to the Charging Current
During charging, charge moves from the power supply to the capacitor, whereas during discharging the stored charge moves from the capacitor into the external circuit. Therefore, the discharge current flows in the opposite direction.
Why the Process Is Considered Complete After About Five Time Constants
An exponential function theoretically never reaches its final value exactly. However, after five time constants, the remaining difference during charging, or the remaining voltage during discharging, is about 0.67% of the initial value.
In many experiments this difference is comparable to or smaller than the measurement error, so charging or discharging is considered practically complete.
Example of How to Write the Results
A circuit with a 10.0 kΩ resistor, 100 μF capacitor, and 5.00 V supply was constructed, and the voltage across the capacitor was measured during charging and discharging.
During charging, the capacitor voltage increased from 0.02 V immediately after the start to 3.14 V at 1.0 s and 4.95 V at 5.0 s. During discharging, it decreased from an initial 4.98 V to 1.86 V at 1.0 s and 0.04 V at 5.0 s.
The time constant obtained from the time at which the charging curve reached 63.2% of its final value was 0.99 s, corresponding to a relative error of 1.0% from the theoretical time constant of 1.00 s. The maximum charge was approximately 500 μC.
Points to Connect to the Discussion
- Were the charging and discharging curves exponential?
- Was the voltage after one time constant close to 63.2% or 36.8%?
- Did the theoretical time constant RC agree with the experimental time constant?
- Did the charge change according to Q=CV?
- Did the charging current decrease with time?
- Was the discharge current direction opposite to the charging direction?
- Did the maximum charge agree with CV0?
- How did increasing resistance affect charging and discharging time?
- How did increasing capacitance affect charging/discharging time and maximum charge?
- Was the semilogarithmic graph linear?
- Did capacitor leakage current or dielectric absorption affect the results?
- Could the input resistance and capacitance of the measuring instrument be neglected?
- Was there error in reading the charging/discharging start time?
Discussion Example
In this experiment, a 10.0 kΩ resistor, a 100 μF capacitor, and a 5.00 V power supply were used to investigate how voltage, charge, and current changed with time during capacitor charging and discharging.
During charging, the capacitor voltage was nearly 0 V immediately after the start, then increased rapidly and approached 5.00 V while the amount of increase gradually became smaller. During discharging, the voltage fell rapidly from its initial value and then approached 0 V more gradually.
These changes approximately agreed with the exponential relations VC = V0{1 − exp(−t/RC)} for charging and VC = V0exp(−t/RC) for discharging.
The theoretical time constant calculated from 10.0 kΩ and 100 μF was 1.00 s. The charging curve reached 3.16 V, corresponding to 63.2% of the final voltage, at about 0.99 s, while the discharge curve fell to 36.8% of its initial voltage at about 1.0 s.
The experimental time constant of 0.99 s was close to the theoretical value of 1.00 s, with a relative error of 1.0%. This result confirms that the charging and discharging rate is determined by the product of resistance and capacitance.
The charge during charging increased according to Q = CV. With a 5.00 V supply and 100 μF capacitance, the maximum charge was calculated as 500 μC, and the experimental value after 5.0 s was about 495 μC.
Immediately after charging began, the capacitor voltage was nearly zero, so almost all the supply voltage appeared across the resistor and the current was maximum. As the capacitor voltage rose, the voltage across the resistor fell, causing the current to decrease exponentially.
After sufficient time, the capacitor voltage approached the supply voltage and the voltage across the resistor became nearly zero. Therefore, in ideal DC steady state the current is nearly zero and the capacitor behaves like an open circuit.
During discharging, the charge stored in the capacitor moved through the resistor into the external circuit, so current flowed in the direction opposite to the charging current. As the capacitor voltage decreased, the magnitude of the current also decreased.
Increasing the resistance reduced the current and therefore reduced the amount of charge transferred per unit time, increasing the time required for charging and discharging. Increasing the capacitance also increased the time constant because more charge was required to reach the same voltage.
Increasing capacitance increased not only the charging and discharging time but also the maximum charge, because Qmax = CV0.
Possible causes of differences between theoretical and experimental values include tolerances in capacitance and resistance, capacitor leakage current, equivalent series resistance, power-supply internal resistance, and measuring-instrument input resistance.
In particular, the input resistance of a voltmeter or oscilloscope is connected in parallel with the capacitor and adds a discharge path. This can make the actual discharge time constant smaller than the theoretical value.
Electrolytic capacitors may also exhibit dielectric absorption, causing the terminal voltage to recover slightly even after sufficient discharge. Therefore, before starting the next charging measurement, the capacitor should be discharged through a resistor for sufficient time.
Near the end of discharging, the voltage is small, so instrument resolution and noise account for a larger fraction of the measured value. Consequently, low-voltage points tend to deviate from a straight line on a semilogarithmic graph, and an appropriate fitting range must be selected.
Measurement accuracy can be improved by measuring the resistance and using the measured value in calculations, fully discharging before charging, charging for at least five time constants before discharging, and taking finer measurements around the time constant.
These results experimentally confirm that capacitor voltage, charge, and current change exponentially during charging and discharging and that the rate of change is determined by the time constant τ = RC.
Discussion Example When the Initial Charging Voltage Was Not Zero
A nonzero capacitor voltage at the start of charging may have been caused by residual charge from the previous measurement or by voltage recovery after discharge due to dielectric absorption. The capacitor should be discharged through a resistor for sufficient time and the initial voltage confirmed before measurement.
Discussion Example When Discharging Was Faster Than the Theoretical Value
Faster-than-theoretical discharging may have resulted from the input resistance of the voltmeter or oscilloscope being connected in parallel with the capacitor and reducing the effective discharge resistance, from capacitor leakage current, or from the actual capacitance being smaller than its nominal value.
Discussion Example When the Final Charging Voltage Was Lower Than the Supply Voltage
Failure of the charged voltage to reach the supply voltage may have resulted from insufficient charging time, a load-induced drop in the power-supply terminal voltage, non-negligible capacitor leakage current, or current division caused by the voltmeter input resistance.
Discussion Example When the Semilogarithmic Graph Was Not Linear
A nonlinear semilogarithmic plot of discharge voltage may have resulted from increased effects of noise and instrument resolution at low voltage, deviation from a simple exponential due to leakage current or dielectric absorption, or inaccurate determination of the discharge start time.
Summary
During capacitor charging and discharging, capacitor voltage, charge, and current change exponentially with time.
The rate of change is determined by the time constant τ = RC. After one time constant, the charging voltage reaches about 63.2% of its final value, while the discharge voltage falls to about 36.8% of its initial value.
In the discussion, it is important to relate the charging and discharging curves, charge, current, maximum charge, stored energy, theoretical and experimental time constants, component tolerances, leakage current, dielectric absorption, and measuring-instrument input resistance to the experimental results.
