Fundamentals of Inductors | Circuit Components

Tutorial 11 — Inductors: V–I–P–E Relationships

An inductor stores energy in its magnetic field. Its terminal behaviour is governed by \(v = L\,\dfrac{di}{dt}\) and, equivalently, \(i(t) = \dfrac{1}{L}\displaystyle\int_{t_0}^{t} v\,d\tau + i(t_0)\), with instantaneous power \(p = vi\) and stored energy \(w = \tfrac{1}{2}L i^2\). These problems work through differentiation of currents, integration of voltage waveforms, energy accounting, equivalent inductance, and DC steady-state behaviour.

Circuit Components · Electric Circuit Analysis · 12 solved problems

Demonstrative VideoWalkthrough

Problem 1Current & Energy

Find the current through a 5 H inductor if the voltage across it is given by

\[ v(t) = \begin{cases} 30t^2, & t > 0 \\ 0, & t < 0 \end{cases} \]

Also find the energy stored at \(t = 5\) s.

Solution

Since \(i = \dfrac{1}{L}\displaystyle\int_{t_0}^{t} v\,d\tau + i(t_0)\) with \(L = 5\) H and \(i(0)=0\):

\[ i = \frac{1}{5}\int_{0}^{t} 30\tau^{2}\,d\tau + 0 = 6 \times \frac{t^{3}}{3} = 2t^{3}\ \text{A} \]

The instantaneous power is \(p = vi = (30t^2)(2t^3) = 60t^5\), so the energy stored is

\[ w = \int_{0}^{5} p\,dt = \int_{0}^{5} 60t^{5}\,dt = \left.60\,\frac{t^{6}}{6}\right|_{0}^{5} = 10\,(5)^6 = 156.25\ \text{kJ} \]

Alternatively, the energy can be obtained directly from the current:

\[ \left.w\right|_{0}^{5} = \tfrac{1}{2}L\,i^{2}(5) - \tfrac{1}{2}L\,i^{2}(0) = \tfrac{1}{2}(5)\left(2\times 5^{3}\right)^{2} - 0 = 156.25\ \text{kJ} \]

The original slide wrote ½L\,i(0) for the second term; it should be ½L\,i^2(0), which is zero here.

Answer\(i(t) = 2t^{3}\ \text{A},\quad w(5) = 156.25\ \text{kJ}\)

Problem 2Voltage & Energy

The current through a 0.1 H inductor is \(i(t) = 10t\,e^{-5t}\) A. Find the voltage across the inductor and the energy stored in it.

Solution

Since \(v = L\,\dfrac{di}{dt}\) with \(L = 0.1\) H:

\[ v = 0.1\,\frac{d}{dt}\!\left(10t\,e^{-5t}\right) = e^{-5t} + t(-5)e^{-5t} = e^{-5t}(1 - 5t)\ \text{V} \]

The energy stored is

\[ w = \tfrac{1}{2}L\,i^{2} = \tfrac{1}{2}(0.1)\,100t^{2}e^{-10t} = 5t^{2}e^{-10t}\ \text{J} \]

Answer\(v(t) = e^{-5t}(1-5t)\ \text{V},\quad w(t) = 5t^{2}e^{-10t}\ \text{J}\)

Problem 3Integration with Initial Condition

The voltage across a 2 H inductor is known to be \(6\cos 5t\) V. Determine the resulting inductor current if \(i(t = -\pi/2) = 1\) A.

Solution

The current through the inductor is

\[ i(t) = \frac{1}{2}\int_{t_0}^{t} 6\cos 5\tau\,d\tau + i(t_0) = \tfrac{1}{2}\!\left(\tfrac{6}{5}\right)\sin 5t - \tfrac{1}{2}\!\left(\tfrac{6}{5}\right)\sin 5t_0 + i(t_0) \]

\[ = 0.6\sin 5t - 0.6\sin 5t_0 + i(t_0) \]

The first term shows the current varies sinusoidally; the remaining terms form a constant fixed by the initial condition. Using \(t_0 = -\pi/2\) with \(i(t_0) = 1\) A and \(\sin(-2.5\pi) = -1\):

\[ i(t) = 0.6\sin 5t - 0.6\sin(-2.5\pi) + 1 = 0.6\sin 5t + 1.6 \]

Answer\(i(t) = 0.6\sin 5t + 1.6\ \text{A}\)

Problem 4Current & Energy

The terminal voltage of a 2 H inductor is \(v = 10(1-t)\) V. Find the current flowing through it at \(t = 4\) s and the energy stored in it at \(t = 4\) s. Assume \(i(0) = 2\) A.

Solution

Given data:

\[ L = 2\ \text{H}, \qquad v(t) = 10(1-t)\ \text{V}, \qquad i(0) = 2\ \text{A} \]

The original slide mistakenly listed L = 2 A; the inductance is 2 H.

Current expression:

\[ i(t) = i(0) + \frac{1}{L}\int_{0}^{t} v\,d\tau = 2 + \frac{1}{2}\int_{0}^{t}(10 - 10\tau)\,d\tau = 2 + 5t - 2.5t^{2} \]

Value of the current at \(t = 4\) s:

\[ i(t)\big|_{t=4} = 2 + 20 - 40 = -18\ \text{A} \]

The energy stored in the inductor at \(t = 4\) s is

\[ w(4) = \tfrac{1}{2}L\,i^{2}(4) = \tfrac{1}{2}(2)(-18)^{2} = 324\ \text{J} \]

The original slide reported 320 J, which is the net energy absorbed over \(0\le t\le 4\), i.e. \(\tfrac12 L\!\left[i^2(4)-i^2(0)\right] = \tfrac12(2)(324-4) = 320\) J. The energy actually stored at \(t=4\) s is \(\tfrac12 L\,i^2(4) = 324\) J.

Answer\(i(4) = -18\ \text{A},\quad w(4) = 324\ \text{J}\ \ (\Delta w_{0\to4} = 320\ \text{J})\)

Problem 5Sawtooth Voltage

If the voltage waveform in the figure is applied to a 10 mH inductor, find the inductor current \(i(t)\). Assume \(i(0) = 0\) A.

Solution

Given \(i(0) = 0\) A and \(L = 10\) mH. The voltage waveform is

\[ v(t) = \begin{cases} 5t, & 0 < t < 1 \\ 5t - 10, & 1 < t < 2 \\ 0, & \text{elsewhere} \end{cases} \]

Integrating \(i = \dfrac{1}{L}\displaystyle\int v\,d\tau\) over each interval (with continuity of current):

\[ i(t) = \begin{cases} 250t^{2}, & 0 < t < 1 \\ 1000\,(1 - t + 0.25t^{2}), & 1 < t < 2 \\ 0, & \text{elsewhere} \end{cases} \]

Check: at \(t=1\), \(250(1)^2 = 250\) A and \(1000(1-1+0.25)=250\) A — the current is continuous; at \(t=2\), \(1000(1-2+1)=0\) A.

Answer\(i_{\max} = i(1) = 250\ \text{A}\), piecewise as above

Problem 6Trapezoidal Current

The current through a 5 mH inductor is shown in the figure. Determine the voltage across the inductor at \(t = 1,\ 3,\ \text{and}\ 5\) ms.

Trapezoidal current waveform for Problem 6

Solution

The current waveform (with \(t\) in ms) is

\[ i = \begin{cases} 5t, & 0 < t < 2\ \text{ms} \\ 10, & 2 < t < 4\ \text{ms} \\ 30 - 5t, & 4 < t < 6\ \text{ms} \end{cases} \]

The slopes are \(\dfrac{di}{dt} = +5000,\ 0,\ -5000\) A/s respectively. Hence \(v = L\dfrac{di}{dt} = 5\times10^{-3}\,\dfrac{di}{dt}\):

\[ v = 5\times10^{-3}\begin{cases} 5000 \\ 0 \\ -5000 \end{cases} = \begin{cases} 25\ \text{V}, & 0 < t < 2\ \text{ms} \\ 0\ \text{V}, & 2 < t < 4\ \text{ms} \\ -25\ \text{V}, & 4 < t < 6\ \text{ms} \end{cases} \]

Note the slope must be expressed in A/s: a rise of 10 A over 2 ms is \(10/0.002 = 5000\) A/s, so \(5\times10^{-3}\times5000 = 25\) V. (The original slide dropped the \(10^{3}\) factor on \(di/dt\).)

Evaluating at the requested instants:

\[ v(1\,\text{ms}) = 25\ \text{V}, \qquad v(3\,\text{ms}) = 0\ \text{V}, \qquad v(5\,\text{ms}) = -25\ \text{V} \]

Answer\(v(1) = 25\ \text{V},\ v(3) = 0\ \text{V},\ v(5) = -25\ \text{V}\)

Problem 7Voltage Waveform & Sketch

Given the waveform of the current in a 3 H inductor as shown in the figure, determine the inductor voltage and sketch it.

Solution

For \(t < -1\) s the current is zero, so the voltage is zero in this interval.

The current then increases at the linear rate of \(1\) A/s, producing a constant voltage

\[ v = L\frac{di}{dt} = 3 \times 1 = 3\ \text{V} \qquad (-1 < t < 0) \]

During the following 2 s the current is constant, so the voltage is zero \((0 < t < 2)\).

The final decrease gives \(\dfrac{di}{dt} = -1\) A/s, yielding \(v = 3\times(-1) = -3\) V for \(2 < t < 3\). For \(t > 3\) s the current is again constant (zero), so \(v(t) = 0\).

Resulting inductor voltage sketch for Problem 7

Answer\(v = 3\ \text{V}\,(-1\!<\!t\!<\!0),\ 0\,(0\!<\!t\!<\!2),\ -3\ \text{V}\,(2\!<\!t\!<\!3)\)

Problem 8Passive Sign Convention

The voltage across a 2 H inductor is given by

\[ v_L = 4.3t, \qquad 0 \le t \le 50\ \text{ms} \]

Knowing that \(i_L(-0.1) = 100\ \mu\text{A}\), calculate the current (defined consistent with the passive sign convention) at

(a) \(t = 0\) ms
(b) \(t = 1.5\) ms
(c) \(t = 45\) ms

Solution

The current through an inductor is

\[ i(t) = \frac{1}{L}\int_{t_0}^{t} v\,d\tau + i_L(t_0) = \frac{1}{2}\int_{-0.1}^{t} 4.3\tau\,d\tau + 100\times10^{-6} = \frac{4.3}{4}\left(t^{2} - 0.01\right) + 100\times10^{-6}\ \text{A} \]

Valid for \(0 \le t \le 50\) ms. (The original slide carried a stray "++" before the \(100\times10^{-6}\) term.)

aAt \(t = 0\):

\[ i(0) = \frac{4.3}{4}\left(0 - 0.01\right) + 100\times10^{-6} = -10.75\times10^{-3} + 100\times10^{-6} = -10.65\ \text{mA} \]

bAt \(t = 1.5\) ms \((= 0.0015\ \text{s})\):

\[ i(0.0015) = \frac{4.3}{4}\left(0.0015^{2} - 0.01\right) + 100\times10^{-6} = -10.748\times10^{-3} + 100\times10^{-6} = -10.648\ \text{mA} \]

cAt \(t = 45\) ms \((= 0.045\ \text{s})\):

\[ i(0.045) = \frac{4.3}{4}\left(0.045^{2} - 0.01\right) + 100\times10^{-6} = -8.573\times10^{-3} + 100\times10^{-6} = -8.473\ \text{mA} \]

Answer\(i(0) = -10.65\ \text{mA},\ i(1.5\,\text{ms}) = -10.648\ \text{mA},\ i(45\,\text{ms}) = -8.473\ \text{mA}\)

Problem 9DC & Sinusoidal Excitation

Determine the current flowing through a 6 mH inductor if the voltage (defined consistent with the passive sign convention) is given by

(a) \(5\) V;
(b) \(100\sin(120\pi t),\ t \ge 0\) and \(0,\ t < 0\).

Solution

The general relation is \(i(t) = \dfrac{1}{L}\displaystyle\int_{0}^{t} v\,d\tau + i_L(0)\), with \(L = 0.006\) H and \(i_L(0) = 0\).

aFor \(v(t) = 5\) V:

\[ i(t) = \frac{1}{0.006}\int_{0}^{t} 5\,d\tau = \frac{1}{0.006}(5t) = 833.33\,t\ \text{A} \]

bFor \(v(t) = 100\sin(120\pi t)\) V:

\[ i(t) = \frac{1}{0.006}\int_{0}^{t} 100\sin(120\pi\tau)\,d\tau = \left.\frac{1}{0.006}\left(-\frac{100}{120\pi}\cos 120\pi\tau\right)\right|_{0}^{t} \]

\[ i(t) = \frac{138.89}{\pi}\bigl(1 - \cos 120\pi t\bigr)\ \text{A} \approx 44.21\bigl(1 - \cos 120\pi t\bigr)\ \text{A}, \qquad t \ge 0 \]

Answer\(i_a = 833.33\,t\ \text{A};\quad i_b = \dfrac{138.89}{\pi}(1-\cos120\pi t)\ \text{A}\)

Problem 10Square-Wave Excitation

A square voltage waveform (amplitude \(\pm 1\) V, period 1 ms) is applied across a 1 mH inductor. Complete the statement: the current is a …… wave of …… peak amplitude.

Solution

The current is the integral of the voltage, \(i_L = \dfrac{1}{L}\displaystyle\int v\,dt\). Writing the square wave with step/ramp functions:

\[ v(t) = u(t) - 2u(t - 0.5) + 2u(t - 1) + \cdots \]

\[ i_L(t) = \frac{1}{L}\bigl[r(t) - 2r(t - 0.5) + 2r(t - 1) + \cdots\bigr] \]

Integrating a constant \(\pm1\) V across \(L = 1\) mH over each 0.5 ms half-period gives a ramp that rises then falls — a triangular current wave. The peak amplitude is

\[ I_p = \frac{1}{L}\,V\,\Delta t = \frac{1}{1\times10^{-3}}\,(1)\,(0.5\times10^{-3}) = 0.5\ \text{A} \]

Resulting triangular current waveform for Problem 10

AnswerThe current is a \(\textbf{triangular}\) wave of \(0.5\ \text{A}\) peak amplitude.

Additional Practice Problems

Two self-contained worked examples with schematics that round out the topic: combining inductors into a single equivalent, and the behaviour of an inductor once a circuit has reached DC steady state.

Problem 11Equivalent Inductance

Find the equivalent inductance \(L_{eq}\) seen at terminals a–b for the network below: a 4 H inductor in series with the parallel combination of 12 H and 6 H, all in series with an 8 H inductor.

Solution

Inductors combine like resistors. First the parallel pair (12 H and 6 H):

\[ L_{12\,\parallel\,6} = \frac{12 \times 6}{12 + 6} = \frac{72}{18} = 4\ \text{H} \]

This is in series with the 4 H and 8 H inductors, so they simply add:

\[ L_{eq} = 4 + 4 + 8 = 16\ \text{H} \]

Answer\(L_{eq} = 16\ \text{H}\)

Problem 12DC Steady State

In the circuit below the source has been connected for a long time, so the circuit is in DC steady state. Find the current \(i_L\) through the 2 H inductor and the energy stored in it. The 6 Ω resistor is connected directly across the inductor.

Solution

At DC steady state the inductor current is constant, so \(v_L = L\,\dfrac{di}{dt} = 0\) — the inductor behaves as a short circuit.

The short across node A–b places 0 V across the parallel 6 Ω resistor, so it carries no current. All the source current therefore flows through the inductor, limited only by the 3 Ω series resistor:

\[ i_L = \frac{V_s}{R_1} = \frac{12}{3} = 4\ \text{A} \]

The energy stored in the inductor is

\[ w = \tfrac{1}{2}L\,i_L^{2} = \tfrac{1}{2}(2)(4)^{2} = 16\ \text{J} \]

Answer\(i_L = 4\ \text{A},\quad w = 16\ \text{J}\)