Capacitor Voltage Formula (Physics Guide)
A capacitor’s voltage is (V_C=Q/C), where (Q) is its charge and (C) is its capacitance. To find how voltage changes over time in a simple resistor-capacitor circuit, use an exponential equation with time constant (\tau=RC). The correct formula depends on the circuit’s starting voltage, final voltage, and switching condition.
A common source of confusion is the idea that a capacitor either “fills up” instantly or must reach its final voltage after a fixed wait. Neither is quite right. In a basic resistor-capacitor, or RC, circuit, voltage changes gradually; it gets very close to its final value but does not reach it in a finite time under the ideal exponential model.
I often see students use a charging equation for a discharging circuit, or treat the source voltage as though it must always equal the capacitor voltage. Choosing the right model comes first. Then the calculation is mostly careful substitution, unit checking, and a quick test to see whether the answer makes sense.
Capacitor Voltage: Formula, Meaning, and When to Use It
A capacitor stores electric charge. Its voltage depends on how much charge it holds and its capacitance. The relation (V_C=Q/C) applies at any instant, but it does not describe how the charge changes over time. For that, use a circuit model that includes current and the surrounding components.
The charge–voltage formula
For a capacitor with charge (Q) and capacitance (C),
[ V_C=\frac{Q}{C} ]
Here, (V_C) is the voltage across the capacitor, measured in volts (V); (Q) is charge, measured in coulombs (C); and (C) is capacitance, measured in farads (F). Rearranging gives (Q=CV_C).
For example, if a (100)-microfarad capacitor holds (0.002) coulombs, then
[ V_C=\frac{0.002\ \text{C}}{100\times10^{-6}\ \text{F}}=20\ \text{V}. ]
The unit conversion matters: (1\ \mu\text{F}=10^{-6}\ \text{F}). This formula tells you the voltage for a known charge. By itself, it cannot tell you how long charging takes.
Voltage, current, and changing charge
Current is the rate at which charge moves. For a capacitor, the current-voltage relation is
[ i_C=C\frac{dV_C}{dt}. ]
In plain language, current causes capacitor voltage to change. If current is constant, the voltage changes at a steady rate. More generally, if you know the current over time, you can find the voltage change by integrating:
[ V_C(t)=V_C(0)+\frac{1}{C}\int_0^t i_C(u)\,du. ]
The sign depends on the chosen current direction. This relation is useful when the current is known, while the RC exponential formula is often easier when a resistor and a switching source set the current.
RC Circuit Curriculum: Charging, Discharging, and Required Parameters
An RC circuit has a resistor and a capacitor. After a switch changes the circuit, the capacitor voltage often follows an exponential curve. To use that curve, identify the starting voltage, the final voltage, and the resistance and capacitance that set the time constant.
Charging from zero voltage
If a DC source (V_s) is connected through a resistor to a capacitor that starts uncharged, then
[ V_C(t)=V_s\left(1-e^{-t/RC}\right). ]
The time (t) is measured in seconds. The quantity (e) is the base of the natural exponential, about 2.718. The product (RC) is the time constant, written as (\tau):
[ \tau=RC. ]
For example, with (R=10\ \text{k}\Omega) and (C=100\ \mu\text{F}),
[ \tau=(10{,}000\ \Omega)(100\times10^{-6}\ \text{F})=1\ \text{s}. ]
If (V_s=12\ \text{V}), then after one second,
[ V_C(1)=12(1-e^{-1})\approx7.6\ \text{V}. ]
That is about (63.2\%) of the source voltage.
Discharging to zero voltage
If a capacitor begins at (V_0) and discharges through a resistor toward zero, use
[ V_C(t)=V_0e^{-t/RC}. ]
Suppose a capacitor starts at (10\ \text{V}), and its time constant is (2\ \text{s}). After (2\ \text{s}), its voltage is about (3.68\ \text{V}), or (36.8\%) of its starting voltage. After (10\ \text{s}), which is (5\tau), it is about (0.067\ \text{V}).
The voltage approaches zero but does not equal zero at any finite time in this ideal equation. In practice, people may call it discharged once it is below a chosen level.
A capacitor with a nonzero starting voltage
Not every capacitor starts at zero, and not every circuit ends at zero. For a simple first-order RC circuit with a constant final voltage (V_f), use
[ V_C(t)=V_f+(V_0-V_f)e^{-t/RC}. ]
Here, (V_0) is the capacitor voltage at the start of the time interval. This general form covers charging, discharging, and a change from one nonzero voltage to another.
For instance, if a capacitor starts at (4\ \text{V}) and moves toward (12\ \text{V}), then after one time constant it has completed about (63.2\%) of the voltage change. The change is (8\ \text{V}), so its voltage is about (4+0.632(8)=9.06\ \text{V}).
Finding the effective resistance
The (R) in (\tau=RC) is the resistance seen by the capacitor during the switching condition being studied. In a simple series circuit, this is the series resistor. In a more complex linear circuit, replace the rest of the network with its Thevenin equivalent: an equivalent voltage source and resistance viewed from the capacitor’s terminals.
Do not automatically use every resistor shown in the diagram, or just the resistor closest to the capacitor. The switch position matters. If the network cannot be reduced to a single first-order RC model, a single exponential may not describe its voltage; use the appropriate circuit equations instead.
Problem-Solving Success: Units, Time Constants, and Common Errors
A reliable solution begins by identifying the capacitor’s initial and final voltages. Next, find the time constant for the circuit after switching, choose the matching equation, and check the result at the start and long after the switch changes. These checks catch many common errors before they spread through the calculation.
Useful time-constant checkpoints
The table gives approximate values for a simple first-order RC circuit. “Percent complete” means the share of the total voltage change completed during charging, or the share of the initial voltage remaining during discharging.
| Elapsed time | Charging from 0 toward (V_s) | Discharging from (V_0) toward 0 |
|---|---|---|
| (1\tau) | (63.2\%) of (V_s) | (36.8\%) of (V_0) remains |
| (3\tau) | (95.0\%) of (V_s) | (5.0\%) of (V_0) remains |
| (5\tau) | (99.3\%) of (V_s) | (0.67\%) of (V_0) remains |
These are checks, not replacement formulas. If a calculated charging voltage falls as time increases, review the sign and initial conditions. If a discharge voltage rises toward zero, check whether you selected the correct equation.
A step-by-step method
- Mark the capacitor voltage. Use (V_C) for voltage across the capacitor, not automatically the source voltage.
- Find (V_0). This is the voltage just after the switching event. In an ordinary ideal circuit, capacitor voltage cannot jump instantly without an impulse of current.
- Find (V_f). Determine the voltage the capacitor approaches in the new circuit condition.
- Find (R) and (C). Use the resistance seen by the capacitor and put capacitance in farads.
- Calculate (\tau=RC). Ohms times farads gives seconds.
- Use the general equation. Substitute (V_0), (V_f), (t), and (\tau).
- Check the limits. At (t=0), the result should equal (V_0). As time grows, it should approach (V_f).
Common mistakes to avoid
- Using (V_C=Q/C) as a time equation. It connects charge and voltage at an instant; it does not specify their change over time.
- Mixing microfarads and farads. Convert before calculating (RC).
- Using the source voltage as the capacitor’s final voltage without checking the circuit. The final value depends on the circuit connections after switching.
- Using a charging formula for discharge. Match the equation to the initial and final conditions.
- Assuming the capacitor reaches its final value exactly at (5\tau). It is close, but the ideal exponential reaches the limit only as time tends toward infinity.
- Ignoring circuit topology. Multiple energy-storage components or changing switch paths may require more than one time constant.
Applications and Career Pathways: Circuits, Electronics, and Engineering Study
Capacitor-voltage equations help describe signals and voltage changes in many circuit problems. The examples below are illustrative calculations, not reports from a particular installation. They show how the same formula applies to different starting and ending voltages, while also showing why the circuit condition must be stated.
Three circuit scenarios
| Scenario | Given conditions | Suitable model | Example result |
|---|---|---|---|
| Charging | (V_s=5\ \text{V}), (R=1\ \text{k}\Omega), (C=100\ \mu\text{F}), (t=0.1\ \text{s}) | (V_s(1-e^{-t/RC})) | (\tau=0.1\ \text{s}); (V_C\approx3.16\ \text{V}) |
| Discharging | (V_0=9\ \text{V}), (R=2\ \text{k}\Omega), (C=50\ \mu\text{F}), (t=0.1\ \text{s}) | (V_0e^{-t/RC}) | (\tau=0.1\ \text{s}); (V_C\approx3.31\ \text{V}) |
| Changing between nonzero levels | (V_0=2\ \text{V}), (V_f=8\ \text{V}), (t=\tau) | (V_f+(V_0-V_f)e^{-t/\tau}) | (V_C\approx5.79\ \text{V}) |
In the third scenario, the capacitor has completed (63.2\%) of the six-volt change: (2+0.632(8-2)\approx5.79\ \text{V}). This “fraction of the change” view is especially useful when the starting voltage is not zero.
When you study this topic, it helps to review charge, current, resistance, capacitance, exponentials, and basic circuit analysis first. If you are planning an engineering program, check the school’s course sequence and whether the specific engineering program is ABET-accredited. Accreditation does not change the physics formula, but it may matter for education and career pathways.
Conclusion: Choose the Formula That Matches the Circuit
Use (V_C=Q/C) when charge and capacitance are known. Use the exponential RC equation when a first-order circuit’s voltage changes after a switching event. The key is to identify the capacitor’s starting and final voltages and the resistance it sees, then confirm the result with units and time-constant checks.
Frequently asked questions
1. What is the basic formula for capacitor voltage?
The basic relation is (V_C=Q/C), where charge is in coulombs and capacitance is in farads.
2. What is the capacitor charging formula?
For charging from zero through a resistor toward a DC source, (V_C(t)=V_s(1-e^{-t/RC})).
3. What is the capacitor discharging formula?
For discharge from (V_0) toward zero through a resistor, (V_C(t)=V_0e^{-t/RC}).
4. What does the time constant mean?
The time constant is (\tau=RC), measured in seconds. After one time constant, a charging capacitor has completed about (63.2\%) of its total voltage change.
5. How do I calculate voltage when the capacitor starts above zero?
Use (V_C(t)=V_f+(V_0-V_f)e^{-t/RC}). Enter the initial voltage (V_0) and the final voltage (V_f) for the circuit after switching.
6. Does a capacitor reach its final voltage after five time constants?
Not exactly in the ideal exponential model. After (5\tau), charging is about (99.3\%) complete, and discharge leaves about (0.67\%) of the starting voltage.
7. What units should I use for (R) and (C)?
Use ohms for resistance and farads for capacitance. Their product gives seconds; convert kilohms and microfarads to base units when needed.
8. Why can’t I use the source voltage as the capacitor voltage?
The source and capacitor voltages can differ during a transient. The capacitor voltage changes over time, and its final value depends on the circuit connections.
9. What if the circuit has several resistors?
Find the equivalent resistance seen by the capacitor in the relevant switch position. For a linear network, this may be found using a Thevenin equivalent.
10. When is one exponential formula not enough?
A single exponential may not work when the circuit has multiple independent energy-storage elements or does not reduce to a first-order RC model. Use the appropriate system of circuit equations.
(This article was written by one of our staff writers, Alan Westbrook. Visit our Meet the Team page.)
