How To Find Voltage Across Capacitor

How to Find Voltage Across a Capacitor

Finding the voltage across a capacitor is a fundamental skill in circuit analysis, whether you are working with direct‑current (DC) sources, alternating‑current (AC) sources, or transient switching events. The voltage across a capacitor tells you how much energy is stored in the electric field between its plates and determines how the component will influence the rest of the network. Below is a step‑by‑step guide that covers the theory, practical methods, and common pitfalls you need to know to determine voltage across capacitor accurately and confidently.

1. Quick Recap: What a Capacitor Does

A capacitor stores charge (Q) proportional to the voltage (V) across its terminals:

[ Q = C , V ]

where (C) is the capacitance (farads). Rearranging gives the basic relationship:

[ V = \frac{Q}{C} ]

Because the voltage cannot change instantaneously in an ideal capacitor (the current (I = C \frac{dV}{dt}) would become infinite), the voltage across a capacitor is a continuous function of time. This property is the key to solving for (V_C(t)) in various circuit conditions.

2. Determining Voltage in DC Steady‑State When a circuit has been connected to a DC source for a long time, all transient effects have died out and the capacitor behaves like an open circuit.

Steps

1. Replace the capacitor with an open circuit (remove it from the schematic).
2. Solve the remaining resistive network using Ohm’s law, Kirchhoff’s voltage law (KVL), or Kirchhoff’s current law (KCL) to find the node voltages. 3. The voltage across the capacitor equals the voltage difference between its two nodes in the opened circuit.

Example

Consider a simple series circuit: a 12 V battery, a 4 kΩ resistor, and a capacitor. After a long time, the capacitor blocks DC current, so the entire battery voltage appears across it:

[ V_C = 12\text{ V} ]

If the capacitor were in parallel with a resistor, the voltage across both would be the same as the source voltage after steady state.

3. Finding Voltage in AC Steady‑State (Phasor Method)

For sinusoidal sources, we use phasor analysis. The capacitor’s impedance is:

[ Z_C = \frac{1}{j\omega C} = -j\frac{1}{\omega C} ]

where (\omega = 2\pi f) is the angular frequency and (j) is the imaginary unit.

Procedure

1. Convert all sources to phasor form (magnitude and angle).
2. Replace each capacitor with its impedance (Z_C).
3. Solve the circuit using standard AC techniques (mesh analysis, nodal analysis, or superposition) treating impedances as complex resistances.
4. The resulting phasor voltage across the capacitor is your answer; convert back to time domain if needed:

[ v_C(t) = |V_C| \cos(\omega t + \angle V_C) ]

Quick Tip

The magnitude of the voltage across a capacitor in a series RC circuit fed by a sinusoidal source (V_s) is:

[ |V_C| = \frac{|V_s|}{\sqrt{1+(\omega RC)^2}} ]

and its phase lags the source by (\phi = -\arctan(\omega RC)).

4. Transient Analysis: Voltage Across a Capacitor During Switching

When a circuit changes state (e.g., a switch closes or opens), the capacitor voltage follows an exponential curve governed by the time constant (\tau = R_{\text{eq}} C).

General Solution

For a first‑order RC circuit with a step input (V_s) applied at (t=0):

[ v_C(t) = V_f + (V_i - V_f) e^{-t/\tau} ]

where

• (V_i) = initial capacitor voltage at (t=0^-)
• (V_f) = final (steady‑state) voltage after the transient has settled
• (\tau) = equivalent resistance seen by the capacitor multiplied by its capacitance

Steps to Apply

1. Find (V_i) – the voltage across the capacitor just before the switching event (often from the previous steady state). 2. Determine the equivalent resistance (R_{\text{eq}}) seen by the capacitor after the switch moves (short voltage sources, open current sources).
2. Compute (\tau = R_{\text{eq}} C). 4. Find (V_f) – the voltage the capacitor would have if left alone for a long time (usually a voltage divider or source value).
3. Plug into the exponential formula to get (v_C(t)) for any time (t\ge0).

Example

A 5 V source, a 2 kΩ resistor, and a 10 µF capacitor are in series. The switch closes at (t=0) connecting the source to the initially uncharged capacitor.

• (V_i = 0) V
• After a long time, the capacitor charges to the source voltage: (V_f = 5) V
• (R_{\text{eq}} = 2) kΩ → (\tau = 2\text{kΩ} \times 10\mu\text{F} = 20\text{ms})

Thus:

[ v_C(t) = 5\bigl(1-e^{-t/20\text{ms}}\bigr)\text{ V} ]

5. Using Kirchhoff’s Laws Directly (Differential Equation Approach)

If the circuit contains multiple capacitors, inductors, or complex sources, writing the governing differential equation can be more systematic.

Procedure

1. Assign a voltage variable (v_C(t)) to each capacitor.
2. Write KCL at each node (or KVL around each loop) expressing currents through capacitors as (i_C = C \frac{dv_C}{dt}).
3. Substitute Ohm’s law for resistors and any source equations. 4. Arrange the equations into a standard linear differential equation form:

[ a_n \frac{d^n v_C}{dt^n} + a_{n-1} \frac{d^{n-1} v_C}{dt^{n-1}} + \dots + a_0 v_C = f(t) ]

5. Solving the Differential Equation

Once the governing differential equation is formulated (as in step 5), solving it requires techniques tailored to its order and complexity. For linear differential equations with constant coefficients, the solution typically involves two parts: the homogeneous solution (response to zero forcing) and the particular solution (response to the input).

Homogeneous Solution

Assume a solution of the form $ v_C(t) = e^{st} $, substitute into the homogeneous equation (set $ f(t) = 0 $), and solve for $ s $ to find the characteristic roots. The roots determine the transient behavior:

• Real and distinct roots: Exponential decay/growth.
• Complex roots: Oscillatory behavior (resonant frequencies).
• Repeated roots: Polynomial terms multiplied by exponentials.

Particular Solution

For non-homogeneous terms (e.g., $ f(t) = V_s \sin(\omega t) $), use methods like undetermined coefficients or variation of parameters to find a steady-state solution matching the input’s form.

Combined Solution

The total solution is the sum of homogeneous and particular solutions:
[ v_C(t) = v_{\text{hom}}(t) + v_{\text{part}}(t) ]
Apply initial conditions (e.g., $ v_C(0) = V_i $) to solve for unknown constants in the homogeneous terms.

Example

For a second-order RLC circuit with a sinusoidal source, the differential equation might resemble:
[ \frac{d^2v_C}{dt^2} + 2\alpha \frac{dv_C}{dt} + \omega_0^2 v_C = V_s \sin(\omega t) ]
Solving this yields both transient (homogeneous) and forced (particular) responses, illustrating how circuits transition to steady-state sinusoidal behavior.

6. Applications and Considerations

Each analysis method applies to

different circuit configurations and provides unique insights. While the nodal analysis method is straightforward for simpler circuits, the differential equation approach offers a more rigorous and versatile framework for complex systems. Specifically, the differential equation approach is invaluable when dealing with circuits containing inductors, which introduce time-dependent behavior not readily captured by nodal analysis. Moreover, it's essential for analyzing circuits exhibiting resonance, where the frequency of the input signal closely matches the natural frequency of the circuit. In these cases, the differential equation approach allows for precise prediction of voltage and current amplitudes and phase shifts.

Beyond fundamental analysis, understanding these methods is crucial for circuit design and optimization. Engineers utilize these techniques to ensure circuits meet specified performance criteria, such as minimizing overshoot, controlling settling time, and maximizing power transfer. Furthermore, the differential equation approach forms the basis for more advanced circuit analysis tools, including simulation software like SPICE. These tools rely on numerically solving differential equations to model circuit behavior with high accuracy.

In conclusion, both nodal analysis and the differential equation approach are powerful tools for circuit analysis. Nodal analysis provides a quick and intuitive method for simpler circuits, while the differential equation approach offers a more rigorous and comprehensive framework for complex systems, particularly those involving inductors and resonant behavior. Mastering both approaches equips engineers with the versatility needed to tackle a wide range of circuit design and analysis challenges. The choice of method depends on the circuit's complexity and the desired level of accuracy. Ultimately, a solid understanding of these techniques is fundamental to the successful design and analysis of electronic systems.