Fundamentals of Capacitors | Circuit Components

Tutorial 13 — Capacitors: V–I–P–E Relationships

A capacitor stores charge in proportion to the voltage across it, \(q = Cv\), and the terminal current follows the rate of change of voltage, \(i = C\,\dfrac{dv}{dt}\). Conversely the voltage is the running integral of current, and the stored energy is \(w = \tfrac{1}{2}Cv^2\). These problems apply those four relations — charge, current, voltage, and energy — together with series/parallel reduction and DC steady-state analysis (where a capacitor behaves as an open circuit).

Circuit Components · Electric Circuit Analysis · 11 solved problems

Demonstrative VideoWalkthrough

Problem 1Charge & Energy

Calculate the charge stored in a 3 pF capacitor with 20 V across it. Also determine the energy stored in the capacitor.

Solution

The stored charge follows directly from \(q = Cv\):

\[ Q = C V = (3\times 10^{-12})(20) = 60\ \text{pC} \]

The energy stored is \(w = \tfrac{1}{2}Cv^2\):

\[ W = \tfrac{1}{2} C V^2 = \tfrac{1}{2}(3\times 10^{-12})(20)^2 = 600\ \text{pJ} \]

Answer\(Q = 60\ \text{pC},\quad W = 600\ \text{pJ}\)

Problem 2Current from Voltage

The voltage across a 5 µF capacitor is \(v(t) = 10\cos 6000t\) V. Calculate the current through it.

Solution

The capacitor current is \(i = C\,\dfrac{dv}{dt}\). Differentiating the cosine introduces a factor of \(-6000\):

\[ i(t) = C\frac{dv}{dt} = (5\times 10^{-6})\frac{d}{dt}\big(10\cos 6000t\big) = -0.3\sin 6000t\ \text{A} \]

Answer\(i(t) = -0.3\sin 6000t\ \text{A}\)

Problem 3Voltage from Current

Determine the voltage across a 2 µF capacitor if the current through it is \(i(t) = 6e^{-3000t}\) mA. Assume the initial capacitor voltage is zero.

Solution

The capacitor voltage is the running integral of the current, with \(v(0)=0\):

\[ v = \frac{1}{C}\int_{0}^{t} i\,dt + v(0) = \frac{1}{2\times 10^{-6}}\int_{0}^{t} 6e^{-3000t}\,dt \cdot 10^{-3} \]

Carrying out the integration:

\[ v = \left.\frac{3\times 10^{3}}{-3000}\,e^{-3000t}\right|_{0}^{t} = \big(1 - e^{-3000t}\big)\ \text{V} \]

Answer\(v(t) = \big(1 - e^{-3000t}\big)\ \text{V}\)

Problem 4Current from Waveform

Determine the current through a 200 µF capacitor whose voltage is shown below.

Triangular voltage waveform across the 200 uF capacitor

Solution

Read the piecewise voltage waveform from the graph:

\[ v(t)=\begin{cases} 50t\ \text{V} & 0\lt t\lt 1 \\ 100-50t\ \text{V} & 1\lt t\lt 3 \\ -200+50t\ \text{V} & 3\lt t\lt 4 \\ 0 & \text{otherwise} \end{cases} \]

Since \(i = C\,dv/dt\) with \(C = 200\ \mu\text{F}\), differentiate each segment (the current is proportional to the slope of \(v\)):

\[ i(t)=200\times 10^{-6}\times\begin{cases} 50 & 0\lt t\lt 1 \\ -50 & 1\lt t\lt 3 \\ 50 & 3\lt t\lt 4 \\ 0 & \text{otherwise} \end{cases} =\begin{cases} 10\ \text{mA} & 0\lt t\lt 1 \\ -10\ \text{mA} & 1\lt t\lt 3 \\ 10\ \text{mA} & 3\lt t\lt 4 \\ 0 & \text{otherwise} \end{cases} \]

Answer\(i = \pm 10\ \text{mA}\) (a square wave, sign following the voltage slope)

Problem 5Energy under DC

Obtain the energy stored in each capacitor under the DC condition.

Resistor network with a 6 mA source and two capacitors

Solution

Under DC steady state, no current flows into a capacitor, so replace each capacitor with an open circuit. The 6 mA source then divides between the 3 kΩ branch and the series \(2\ \text{k}\Omega + 4\ \text{k}\Omega\) branch. The current in the series branch is found by current division:

\[ i = \frac{3}{3+2+4}\,(6\ \text{mA}) = 2\ \text{mA} \]

The voltage across each capacitor equals the voltage across the resistor it shunts:

\[ v_1 = 2000\,i = 4\ \text{V},\qquad v_2 = 4000\,i = 8\ \text{V} \]

The stored energy in each capacitor, \(w = \tfrac{1}{2}Cv^2\):

\[ \begin{aligned} w_1 &= \tfrac{1}{2}C_1 v_1^{2} = \tfrac{1}{2}(2\times 10^{-3})(4)^{2} = 16\ \text{mJ}\\ w_2 &= \tfrac{1}{2}C_2 v_2^{2} = \tfrac{1}{2}(4\times 10^{-3})(8)^{2} = 128\ \text{mJ} \end{aligned} \]

Answer\(w_1 = 16\ \text{mJ},\quad w_2 = 128\ \text{mJ}\)

Problem 6Equivalent Capacitance

Find the equivalent capacitance between the terminals.

Network of five capacitors between two terminals

Solution

Combine the series 20 µF and 5 µF capacitors first (series capacitors add reciprocally):

\[ \frac{20\times 5}{20+5} = 4\ \mu\text{F} \]

This 4 µF is now in parallel with the 6 µF and 20 µF capacitors (parallel capacitors add directly):

\[ 4 + 6 + 20 = 30\ \mu\text{F} \]

Finally, this 30 µF is in series with the 60 µF capacitor, giving the equivalent capacitance of the whole network:

\[ C_{\text{eq}} = \frac{30\times 60}{30+60} = 20\ \mu\text{F} \]

Answer\(C_{\text{eq}} = 20\ \mu\text{F}\)

Problem 7Voltage Division (Series Caps)

For the network below, a 30 V source drives \(C_1 = 20\ \text{mF}\) and \(C_2 = 30\ \text{mF}\) in series with a parallel pair (\(40\ \text{mF}\,\|\,20\ \text{mF}\)). Find the voltage across each capacitor (\(v_1\), \(v_2\), and \(v_3\) across the parallel pair).

Note: the worked values follow the standard textbook problem (Sadiku, Ex. 6.10). The slide image used µF labels, but the consistent set that satisfies the 30 V loop is the millifarad set shown above.

Solution

First find the equivalent capacitance. The two parallel capacitors combine to:

\[ 40 + 20 = 60\ \text{mF} \]

This 60 mF is in series with the 20 mF and 30 mF capacitors, so the three series elements give:

\[ C_{\text{eq}} = \frac{1}{\frac{1}{60}+\frac{1}{30}+\frac{1}{20}}\ \text{mF} = 10\ \text{mF} \]

The total charge delivered by the 30 V source is the same charge carried by every series element:

\[ q = C_{\text{eq}}\,v = (10\times 10^{-3})(30) = 0.3\ \text{C} \]

Because \(C_1\) (20 mF) and \(C_2\) (30 mF) are in series with the source, each carries this 0.3 C. Their voltages are:

\[ v_1 = \frac{q}{C_1} = \frac{0.3}{20\times 10^{-3}} = 15\ \text{V},\qquad v_2 = \frac{q}{C_2} = \frac{0.3}{30\times 10^{-3}} = 10\ \text{V} \]

Finally, the voltage across the parallel pair follows from KVL (or equivalently from \(q/C\) on the 60 mF combination):

\[ v_3 = 30 - v_1 - v_2 = 5\ \text{V} = \frac{q}{60\ \text{mF}} = \frac{0.3}{60\times 10^{-3}} = 5\ \text{V} \]

Answer\(v_1 = 15\ \text{V},\quad v_2 = 10\ \text{V},\quad v_3 = 5\ \text{V}\)

Problem 8Voltage from Source Current

Assuming the capacitors are initially uncharged, find \(v_0\). The 6 µF and 3 µF capacitors are in series across the source current, and \(v_0\) is taken across the 3 µF capacitor.

Triangular source current feeding 6 uF and 3 uF capacitors

Solution

The voltage across the 3 µF capacitor is the integral of the source current, starting from \(v(0)=0\):

\[ v_0 = \frac{1}{C}\int_{0}^{t} i\,dt + v(0) \]

For \(0\lt t\lt 1\), the source ramps up as \(i = 90t\) mA:

\[ v_0 = \frac{10^{-3}}{3\times 10^{-6}}\int_{0}^{t} 90t\,dt + 0 = 15t^{2}\ \text{kV},\qquad v_0(1) = 15\ \text{kV} \]

For \(1\lt t\lt 2\), the source ramps down as \(i = (180-90t)\) mA. Integrating from \(t=1\) and adding the value \(v_0(1)\):

\[ \begin{aligned} v_0 &= \frac{10^{-3}}{3\times 10^{-6}}\int_{1}^{t}(180-90t)\,dt + v_0(1)\\ &= \big[60t - 15t^{2}\big]_{1}^{t} + 15\ \text{kV}\\ &= \big[60t - 15t^{2} - (60-15) + 15\big]\ \text{kV} = \big[60t - 15t^{2} - 30\big]\ \text{kV} \end{aligned} \]

Collecting both intervals:

\[ v_0(t)=\begin{cases} 15t^{2}\ \text{kV}, & 0\lt t\lt 1\\ \big[60t - 15t^{2} - 30\big]\ \text{kV}, & 1\lt t\lt 2 \end{cases} \]

Answer\(v_0(1) = 15\ \text{kV}\); piecewise as above

Problem 9Parallel Caps — v, i₁, i₂

If \(v(0)=0\), find \(v(t)\), \(i_1(t)\) and \(i_2(t)\) for the 6 µF and 4 µF capacitors driven by the source current shown.

Source-current waveform feeding parallel 6 uF and 4 uF capacitors

Solution

Read the piecewise source current, and note the two capacitors are in parallel so they share the same voltage with \(C_{\text{eq}} = 6 + 4 = 10\ \mu\text{F}\):

\[ i_s(t)=\begin{cases} 30t\ \text{mA}, & 0\lt t\lt 1\\ 30\ \text{mA}, & 1\lt t\lt 3\\ (-75+15t)\ \text{mA}, & 3\lt t\lt 5 \end{cases} \qquad v = \frac{1}{C_{\text{eq}}}\int_{0}^{t} i\,dt + v(0) \]

For \(0\lt t\lt 1\) (with \(i = 30t\) mA):

\[ v = \frac{10^{-3}}{10\times 10^{-6}}\int_{0}^{t} 30t\,dt + 0 = 1.5t^{2}\ \text{kV} \]

For \(1\lt t\lt 3\) (with \(i = 30\) mA), integrating from \(t=1\) and adding \(v(1)=1.5\) kV:

\[ v = \frac{10^{3}}{10}\int_{1}^{t} 30\,dt + v(1) = \big[3(t-1)+1.5\big]\ \text{kV} = \big[3t-1.5\big]\ \text{kV} \]

For \(3\lt t\lt 5\) (with \(i = 15(t-5)\) mA), integrating from \(t=3\) and adding \(v(3)=7.5\) kV:

\[ \begin{aligned} v &= \frac{10^{3}}{10}\int_{3}^{t} 15(t-5)\,dt + v(3)\\ &= \left[0.75t^{2} - 7.5t\right]_{3}^{t} + 7.5\ \text{kV} = \big[0.75t^{2} - 7.5t + 23.25\big]\ \text{kV} \end{aligned} \]

Collecting the voltage:

\[ v(t)=\begin{cases} 1.5t^{2}\ \text{kV}, & 0\lt t\lt 1\,\text{s}\\ \big[3t-1.5\big]\ \text{kV}, & 1\lt t\lt 3\,\text{s}\\ \big[0.75t^{2}-7.5t+23.25\big]\ \text{kV}, & 3\lt t\lt 5\,\text{s} \end{cases} \]

Each branch current is \(i_k = C_k\,dv/dt\). With \(C_1 = 6\ \mu\text{F}\):

\[ i_1 = 6\times 10^{-6}\frac{dv}{dt} = \begin{cases} 18t\ \text{mA}, & 0\lt t\lt 1\,\text{s}\\ 18\ \text{mA}, & 1\lt t\lt 3\,\text{s}\\ (9t-45)\ \text{mA}, & 3\lt t\lt 5\,\text{s} \end{cases} \]

With \(C_2 = 4\ \mu\text{F}\):

\[ i_2 = 4\times 10^{-6}\frac{dv}{dt} = \begin{cases} 12t\ \text{mA}, & 0\lt t\lt 1\,\text{s}\\ 12\ \text{mA}, & 1\lt t\lt 3\,\text{s}\\ (6t-30)\ \text{mA}, & 3\lt t\lt 5\,\text{s} \end{cases} \]

Answer\(i_1:i_2 = 6:4\) at every instant, as listed piecewise above

Additional Practice Problems

Two self-contained worked examples that round out the V–I–P–E picture: computing capacitance from the physical geometry of a parallel-plate capacitor, and the classic charge-redistribution problem when a charged capacitor is connected to an uncharged one.

Problem 10Parallel-Plate Capacitor

A parallel-plate capacitor has plate area \(A = 100\ \text{cm}^2\) and plate separation \(d = 0.1\ \text{mm}\), filled with mica (\(\varepsilon_r = 6\)). Find its capacitance, and the charge and energy stored when connected to 100 V. Use \(\varepsilon_0 = 8.854\times 10^{-12}\ \text{F/m}\).

Solution

Convert to SI units: \(A = 100\ \text{cm}^2 = 10^{-2}\ \text{m}^2\) and \(d = 0.1\ \text{mm} = 10^{-4}\ \text{m}\). The parallel-plate capacitance is:

\[ C = \frac{\varepsilon_0 \varepsilon_r A}{d} = \frac{(8.854\times 10^{-12})(6)(10^{-2})}{10^{-4}} = 5.31\ \text{nF} \]

The charge stored at 100 V:

\[ Q = CV = (5.31\times 10^{-9})(100) = 531\ \text{nC} \]

The energy stored:

\[ W = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(5.31\times 10^{-9})(100)^2 = 26.6\ \mu\text{J} \]

Answer\(C = 5.31\ \text{nF},\quad Q = 531\ \text{nC},\quad W = 26.6\ \mu\text{J}\)

Problem 11Charge Redistribution

A 4 µF capacitor is charged to 50 V and then connected (through a switch) across an uncharged 6 µF capacitor. Find the common voltage after the switch closes and the energy lost in the process.

Solution

Charge is conserved when the switch closes (no source is present). The initial charge resides entirely on \(C_1\):

\[ Q = C_1 V_0 = (4\times 10^{-6})(50) = 200\ \mu\text{C} \]

After closing, the two capacitors are in parallel and share that charge at a common voltage:

\[ V_f = \frac{Q}{C_1 + C_2} = \frac{200\ \mu\text{C}}{(4+6)\ \mu\text{F}} = 20\ \text{V} \]

Compare stored energy before and after:

\[ \begin{aligned} W_i &= \tfrac{1}{2}C_1 V_0^2 = \tfrac{1}{2}(4\times 10^{-6})(50)^2 = 5\ \text{mJ}\\ W_f &= \tfrac{1}{2}(C_1+C_2)V_f^2 = \tfrac{1}{2}(10\times 10^{-6})(20)^2 = 2\ \text{mJ} \end{aligned} \]

The difference is dissipated (in the wire resistance and the connecting arc), regardless of how small the resistance is:

\[ \Delta W = W_i - W_f = 5 - 2 = 3\ \text{mJ} \]

Answer\(V_f = 20\ \text{V},\quad \Delta W = 3\ \text{mJ lost}\)