Alternating Current: Differential Equation Approach

Before moving to phasor analysis of resistive, capacitive, and inductive circuits, this chapter looks at analysis of such circuits using differential equations directly. The aim is to show that phasor analysis makes our lives much easier.

For an excellent review of the mathematics of solving linear, first order, constant coefficient differential equations, see Dawkins [2022].

Voltage Divider

Example

Find v\(_O\)(t) given that \(v_I(t)=4 \cos(10000t+45^\circ)\) V

Solution using differential equations

First remember that

\[v_O(t) = L \frac{d~i(t)}{dt}\]

and that

\[ v_R(t) = R i(t).\]

Then, from Kirchhoff’s Voltage Law:

\[ v_I(t) = v_R(t) + v_O(t)\]

or

\[ 4 \cos(10000t+45^\circ) = R i(t) + L \frac{d~i(t)}{dt}\]

Now, we need to solve this linear, first order differential equation (Dawkins [2022]).

Note

Dawkins [2022] says to:

• Put the differential equation in the correct initial form.

\[ \frac{d~i(t)}{dt} + p(t) i(t) = g(t) \]

• Find the integrating factor \(\mu(t)\) using

\[ \mu(t) = e^{\int p(t) dt} \]

• Multiply everything in the differential equation by \(\mu(t)\) and verify that the left side becomes the product rule \((\mu(t)y(t))^′\) and write it as such.

• Integrate both sides, make sure you properly deal with the constant of integration.

• Solve for the solution \(i(t)\).

To get the equation into the correct form, just swap sides and divide both sides by \(L\):

\[ \frac{R}{L} i(t) + \frac{d~i(t)}{dt} = \frac{4}{L} \cos(10000t+45^\circ)\]

Then we can see that \(p(t) = \frac{R}{L}\) so that

\[ \mu(t) = e^{\frac{R}{L} t} \]

Then

\[\begin{align*} \mu(t)\frac{R}{L} i(t) + \mu(t)\frac{d~i(t)}{dt} &= \mu(t)\frac{4}{L} \cos(10000t+45^\circ)\\ (\mu(t) i(t))' &= \mu(t)\frac{4}{L} \cos(10000t+45^\circ) \end{align*}\]

and integrating both sides we get

\[\begin{align*} \int (e^{\frac{R}{L} t} i(t))' dt &= \int e^{\frac{R}{L} t}\frac{4}{L} \cos(10000t+45^\circ) dt\\ e^{\frac{R}{L} t} i(t) + k &= \frac{4 e^{\frac{R}{L} t}}{10^8 L^2 + R^2} \big [ 10000 L \sin(10000t+45^\circ) \\ &+ R \cos(10000t+45^\circ) \big ] + c \end{align*}\]

so that

\[\begin{align*} i(t) &= \frac{4}{10^8 L^2 + R^2} \big [ 10000 L \sin(10000t+45^\circ) \\ &+ R \cos(10000t+45^\circ) \big ] \\ & + \kappa e^{-\frac{R}{L} t} \end{align*}\]

Substituting \(L= 0.1\) and \(R = 1000\) we get

\[\begin{align*} i(t) &= \frac{4}{10^6 + 10^6} \big [ 1000 \sin(10000t+45^\circ)\\ &+ 1000 \cos(10000t+45^\circ) \big ] \\ &+ \kappa e^{-\frac{R}{L} t} \\ &= \frac{4}{10^6 + 10^6} 1000 \sqrt{2} \cos(10000 t) + \kappa e^{-\frac{R}{L} t} \end{align*}\]

And after a long time (\(t > 5 \frac{L}{R} = 500 \mu s\))

\[ i(t) = \frac{2 \sqrt{2}}{1000} \cos(10000 t) \]

So

\[\begin{align*} v_O(t) &= L \frac{d~i(t)}{dt} \\ &= 0.1 \times ( - 10000 \times \frac{2 \sqrt{2}}{1000} \sin(10000t) )\\ &= - 2 \sqrt{2} \sin(10000t) \\ &= 2 \sqrt{2} \cos(10000 t + 90^\circ) \end{align*}\]

Current Divider

Example

Find \(i_O(t)\) given that \(i_I(t)=400 \cos(1000t-30^\circ)\) mA.

Solution using differential equations

Here

\[ i_O(t) = C \frac{d~v_C(t)}{dt} \]

and applying Kirchhoff’s Current Law (KCL) to the top node we get

\[ i_I(t) = v_R(t) / R + C \frac{d~v_C(t)}{dt} \]

Rearranging this into Dawkins’ preferred format, we get

\[ \frac{1}{RC} v_R(t) + \frac{d~v_C(t)}{dt} = 400 \cos(1000t-30^\circ) \]

and noting that \(v_R(t) = v_C(t)\)

\[ \frac{1}{RC} v(t) + \frac{d~v(t)}{dt} = 400 \cos(1000t-30^\circ) \]

For this example, our integrating factor \(\mu(t)\) is now

\[ \mu(t) = e^{\frac{t}{RC}} \]

and multiplying both sides of our differential equation by this factor yields

\[ \frac{d}{dt} (\mu(t) v(t)) = \mu(t) 400 \cos(1000t-30^\circ ).\]

We can now integrate this as we did in the voltage divider example to give

\[\begin{align*} \int \frac{d}{dt} (e^{\frac{t}{RC}} v(t)) dt &= \int e^{\frac{t}{RC}} 400 \cos(1000t-30^\circ) dt \\ \int \frac{d}{dt} (e^{\frac{t}{5 \times 10^{-4}}} v(t)) dt &= \int e^{\frac{t}{5 \times 10^{-4}}} 400 \cos(1000t-30^\circ) dt \\ e^{2000 t} v(t) + c &= 0.178885 e^{2000 t} \sin(1000 t + 33.4^\circ ) + k\\ v(t) &= 0.178885 \sin(1000 t + 33.4^\circ ) + \kappa e^{-2000 t} \end{align*}\]

And after a long time (\(t > 5 RC = 2.5 ms\)):

\[ v(t) = 0.178885 \sin(1000 t + 33.4^\circ ) \]

Then

\[\begin{align*} i_O(t) &= C \frac{d~v(t)}{dt} \\ &= 1 \times 10^{-6} \times ( 178.885 \cos(1000 t + 33.4^\circ) )\\ &= 178.885 \cos(1000 t + 33.4^\circ) \mu A \end{align*}\]

Mesh Analysis

Example

Solve for \(v_O(t)\) using mesh analysis where

\[ v_S(t) = 12 \cos( 100 t) \mbox{V} \]

and

\[ i_S(t) = 4 \cos( 100 t + 90^\circ) \mbox{A}. \]

Solution using differential equations

First, let’s define the mesh currents \(I_1\) and \(I_2\).

\(I_1\) Mesh

Then, the KVL equation for the \(I_1\) mesh is

\[\begin{align*} -v_s(t) + v_{R1}(t) + v_C(t) + v_{R2}(t) &= 0\\ -12\cos( 100 t) + I_1(t) R_1 + \ldots \\ \frac{1}{C} \int I_1(t) + I_2(t) dt + (I_1(t) + I_2(t)) R_2 &= 0\\ -12\cos( 100 t) + 4 I_1(t) + \ldots \\ 1000 \int I_1(t) + I_2(t) dt + 5(I_1(t) + I_2(t)) &= 0 \end{align*}\]

yielding

(19)\[9 I_1(t) + 1000 \int I_1(t) + I_2(t) dt + 5 I_2(t) = 12\cos( 100 t).\]

\(I_2\) Mesh

Looking at the \(I_2\) mesh, we see that there is a current source and that \(I_2\) is in the same direction as \(i_S\).

This simplifies the analysis because it means

\[ I_2(t) = i_S(t) = 4 \cos( 100 t + 90^\circ) \mbox{A}. \]

Combining the \(I_1\) and \(I_2\) equations

Substituting this expression for \(I_2\) into (19) gives

\[\begin{align*} 9 I_1(t) + 1000 \int I_1(t) &+ 4 \cos( 100 t + 90^\circ) dt \\ &+ 20 \cos( 100 t + 90^\circ) = 12\cos( 100 t)\\ 9 I_1(t) + 1000 \int I_1(t) &+ 4 \cos( 100 t + 90^\circ) dt\\ &= 12\cos( 100 t) - 20 \cos( 100 t + 90^\circ)\\ 9 I_1(t) + 1000 \int I_1(t) dt & \\ &= 23.32 \cos(100 t - 59.03^\circ) \\ &- 4000 \int \cos( 100 t + 90^\circ) dt \end{align*}\]

Then, to turn this into a differential equation, we can differentiate both sides with respect to \(t\) to get:

\[\begin{align*} 9 \frac{d}{dt} I_1(t) + 1000 I_1(t) &= -2332 \sin(100 t - 59.03^\circ)\\ &-4000 \cos( 100 t + 90^\circ) \\ \frac{d}{dt} I_1(t) + 111.111 I_1(t) &= 259.11 \cos(100 t + 30.97^\circ)\\ &-444.444 \cos( 100 t + 90^\circ) \\ \frac{d}{dt} I_1(t) + 111.111 I_1(t) &= 382.294 \cos(100 t -54.47^\circ)\\ \end{align*}\]

Again using our integrating factor \(\mu(t) = e^{t/0.009}\), we get

\[ \frac{d}{dt} (\mu(t) I_1(t)) = 382.294 \cos(100 t -54.47^\circ) \mu(t). \]

Integrating both sides with respect to \(t\) gives

\[\begin{align*} e^{t/0.009} I_1(t) + c &= 382.294 \int e^{t/0.009} \cos(100 t -54.47^\circ) dt \\ e^{t/0.009} I_1(t) + c &= -2.55741 e^{t/0.009} \sin(100 t + 173.54^\circ) + k \end{align*}\]

Rearranging, we get

\[\begin{align*} I_1(t) &= -2.55741 e^{t/0.009} \sin(100 t + 173.54^\circ) + \kappa e^{-t/0.009}\\ &= 2.55741 \cos(100t + 263.54^\circ) + \kappa e^{-t/0.009} \end{align*}\]

and then after more than \(t \gg 5 \times 0.009\) we get

\[\begin{align*} I_1(t) &= 2.55741 \cos(100t + 263.54^\circ) \mbox{A.} \end{align*}\]

Solving for \(v_O(t)\)

Then

\[\begin{align*} v_O(t) &= (I_1(t) + I_2(t)) R_2 \\ &= 12.78705 \cos(100t + 263.54^\circ) \\ &+ 20 \cos( 100 t + 90^\circ) \\ &= 7.4347 \cos(100 t + 101.16^\circ) \mbox{V.} \end{align*}\]

Nodal Analysis

Example

Find \(v_O(t)\) using nodal analysis where

\[ v_S(t) = 6 \cos(100t) \mbox{V} \]

and

\[ i_S(t) = 4 \cos(100 t + 45^\circ) \mbox{A.} \]

Solution using differential equations

First, let’s define the nodes and the component current directions.

Inductor Current

The current in inductor \(L\) can be derived from

\[ v_L(t) = L \frac{d}{dt} i_L(t) \]

so

\[\begin{align*} i_L(t) &= \frac{1}{L} \int v_L(t) dt + c\\ &= \frac{1}{L} \int \left(v_X(t) - v_Y(t)\right) dt + c \end{align*}\]

which is in terms of the node voltages.

Capacitor Current

The current in the capacitor \(C\) is

\[ i_C(t) = C \frac{d}{dt} v_C(t) = C \frac{d}{dt} \left( v_Y(t) - v_Z(t) \right) \]

Resistor Current

The current in the resistor \(R\) is

\[ i_R(t) = v_Z(t)/R \]

Node \(\color{red}{\bf X}\) :

At node \(\color{red}{\text{X}}\) we can immediately see that

\[ v_X(t) = v_S(t) = 6 \cos(100t) \]

Node \(\color{green}{\bf Y}\) :

At node \(\color{green}{\text{Y}}\), the KCL equation is

\[\begin{align*} i_L(t) - i_S(t) - i_C(t) &= 0\\ \frac{1}{L} \int \left(v_X(t) - v_Y(t)\right) dt + c &\\ - 4 \cos(100 t + 45^\circ) - C \frac{d}{dt} \left( v_Y(t) - v_Z(t) \right) &= 0\\ \frac{1}{L} \int \left(6 \cos(100t) - v_Y(t)\right) dt + c &\\ - 4 \cos(100 t + 45^\circ) - C \frac{d}{dt} \left( v_Y(t) - v_Z(t) \right) &= 0 \end{align*}\]

Taking the derivative with respect to \(t\) gives

(20)\[\begin{split}&\frac{6}{L} \cos(100t) - \frac{1}{L} v_Y(t)\\ &+ 400 \sin(100 t + 45^\circ)\\ &- C \frac{d^2}{dt^2} \left( v_Y(t) - v_Z(t) \right) = 0\end{split}\]

Node \(\color{blue}{\bf Z}\) :

At node \(\color{blue}{\text{Z}}\), the KCL equation is

\[\begin{align*} i_C(t) - i_R(t) &= 0\\ C \frac{d}{dt} \left( v_Y(t) - v_Z(t) \right) - v_Z(t)/R &= 0\\ RC \frac{d}{dt} \left( v_Y(t) - v_Z(t) \right) - v_Z(t) &= 0 \end{align*}\]

Then, integrating the equation gives:

\[\begin{align*} RC \left( v_Y(t) - v_Z(t) \right) - \int v_Z(t) dt + c &= 0 \end{align*}\]

(21)\[v_Y(t) = v_Z(t) + \frac{1}{RC} \int v_Z(t) dt + c\]

Substituting

Substitution of (21) into (20) then yields:

\[\begin{align*} &\frac{6}{L} \cos(100t) \\ &- \frac{1}{L} ( v_Z(t) + \frac{1}{RC} \int v_Z(t) dt + c)\\ & + 400 \sin(100 t + 45^\circ) \\ &- C \frac{d^2}{dt^2} \left(v_Z(t) + \frac{1}{RC} \int v_Z(t) dt + c - v_Z(t) \right) \\ &= 0 \end{align*}\]

which gives

\[\begin{align*} &\frac{6}{L} \cos(100t) \\ &- \frac{1}{L} v_Z(t) - \frac{1}{RCL} \int v_Z(t) dt - \frac{c}{L} \\ & + 400 \sin(100 t + 45^\circ) \\ &- \frac{d^2}{dt^2} \left( \frac{1}{R} \int v_Z(t) dt + c \right) \\ &= 0 \end{align*}\]

or

\[\begin{align*} &\frac{6}{L} \cos(100t) \\ &- \frac{1}{L} v_Z(t) - \frac{1}{RCL} \int v_Z(t) dt - \frac{c}{L} \\ & + 400 \sin(100 t + 45^\circ) \\ &- \frac{1}{R} \frac{d v_Z(t)}{dt} \\ &= 0 \end{align*}\]

Taking the derivative of the equation with respect to \(t\) yields

\[\begin{align*} &-\frac{600}{L} \sin(100t) \\ &- \frac{1}{L} \frac{d v_Z(t)}{dt} - \frac{1}{RCL} v_Z(t) \\ & + 40000 \cos(100 t + 45^\circ) \\ &- \frac{1}{R} \frac{d^2 v_Z(t)}{dt^2} \\ &= 0\\ & 0.01 \frac{d^2 v_Z(t)}{dt^2} - 10 \frac{d v_Z(t)}{dt} - 1000 v_Z(t) = \\ &6000 \sin(100t) - 40000 \cos(100 t + 45^\circ)\\ & 0.01 \frac{d^2 v_Z(t)}{dt^2} - 10 \frac{d v_Z(t)}{dt} - 1000 v_Z(t) = \\ &-6000 \cos(100t + 90^\circ) - 40000 \cos(100 t + 45^\circ)\\ & 0.01 \frac{d^2 v_Z(t)}{dt^2} - 10 \frac{d v_Z(t)}{dt} - 1000 v_Z(t) = \\ & 44446 \cos(100 t - 129.5^\circ) \end{align*}\]

We can then solve this using Dawkins [2022] to find the homogenous and particular solutions.

Homogenous Solution

The characteristic equation for the homogenous solutions is

\[ 0.01 r^2 - 10 r - 1000 = 0\]

which has solutions

\[\begin{align*} r &= \frac{ 10 \pm \sqrt{100 - 4 \times 0.01 \times -1000}}{2 \times 0.01}\\ &= 1091.6 \mbox{ or } -91.6 \end{align*}\]

So the homogenous solution is

\[\begin{align*} v_{Z_h}(t) = c_1 e^{1091.6 t} + c_2 e^{-91.6 t} \end{align*}\]

where \(c_1\) and \(c_2\) are constants that need to be determined by initial or other conditions on \(v_{Z_h}(t)\).

Particular Solution

To find the particular solution, assume

\[ v_{Z_p}(t) = A \cos(100 t + B) \]

Substituting this into the left-hand side of the differential equation gives

\[\begin{align*} &- 0.01 \times 100^2 A \cos(100 t + B)\\ &- 10 \times (-100 A \sin( 100 t + B))\\ & - 1000 A \cos(100 t + B)\\ &=\\ &- 100 A \cos(100 t + B)\\ &+ 1000 A \sin( 100 t + B)\\ & - 1000 A \cos(100 t + B)\\ &=\\ &- 100 A \cos(100 t + B)\\ &- 1000 A \cos( 100 t + B + 90^\circ)\\ & - 1000 A \cos(100 t + B)\\ &= 1486 A \cos(100 t + B - 137^\circ) \end{align*}\]

Equating the two sides of the equation

\[\begin{align*} & 1486 A \cos(100 t + B - 137^\circ) \\ & = 44446 \cos(100 t - 129.5^\circ) \end{align*}\]

shows that

\[ A = 44446/1486 = 29.91 \]

and

\[ B = 7.5^\circ \]

so that

\[ v_{Z_p}(t) = 29.91 \cos(100 t + 7.5^\circ) \mbox{V} \]

Finding \(v_Y(t)\)

Then we can substitute \(v_{Z_p}(t)\) into (21) to get

\[\begin{align*} v_Y(t) &= 29.91 \cos(100 t + 7.5^\circ) + \frac{1}{RC} \int 29.91 \cos(100 t + 7.5^\circ) dt + c\\ &= 29.91 \cos(100 t + 7.5^\circ) + 100 \int 29.91 \cos(100 t + 7.5^\circ) dt + c\\ &= 29.91 \cos(100 t + 7.5^\circ) + 29.91 \sin(100 t + 7.5^\circ) + c\\ &= 29.91 \cos(100 t + 7.5^\circ) - 29.91 \cos(100 t + 97.5^\circ) + c\\ &= 42.299 \cos(100 t - 37.5^\circ) \end{align*}\]

Finding \(v_O(t)\)

Then

\[ v_O(t) = v_Z(t) = 29.91 \cos(100 t + 7.5^\circ) \mbox{V}. \]

References

Daw22a(1,2,3)

Paul Dawkins. Differential Equations - Linear Equations. https://tutorial.math.lamar.edu/classes/de/Linear.aspx, 2022. [Online; accessed 14-March-2023].

Daw22b

Paul Dawkins. Differential Equations - Second Order DEs. https://tutorial.math.lamar.edu/Classes/DE/SecondOrderConcepts.aspx, 2022. [Online; accessed 28-June-2023].