Circuit Equations

There are two (related) approaches:

1. Derive the circuit (differential) equations in the time domain, then transform these ODEs to the s-domain;
2. Transform the circuit to the s-domain, then derive the circuit equations in the s-domain (using the concept of "impedance").

We will use the first approach. We will derive the system equations(s) in the t-plane, then transform the equations to the s-plane. We will usually then transform back to the t-plane.

EXAMPLE 1

Consider the circuit when the switch is closed at t = 0 with V_C(0) = 1.0 V. Solve for the current i(t) in the circuit.

Answer

Multiplying throughout by 10⁻⁶:

Taking Laplace transform:


Now in this example, we are told .

So

That is: Therefore:
NOTE:


Collecting I terms and subtracting from both sides:


Finding the inverse Laplace transform gives us

Note: Throughout this page these problems are also solved using Scientific Notebook. They are TEX files and you need Scientific Notebook or similar, to view them.

Alternative answer using Scientific Notebook. (.tex file)

EXAMPLE 2

Solve for i(t) for the circuit, given that V(t) = 10 sin5t V, R = 4 W and L = 2 H.

Answer

So


25 = A(s² + 25) + (Bs + C)(s + 2)
We need to solve for A, B and C.
First, let s = −2 and this gives
25 = 29A
Thus A = 25/29
Next, we equate coefficients of s²:
gives
Equating coefficients of :
gives
So

So we have
A

Alternative answer using Scientific Notebook. (.tex file)

EXAMPLE 3

In the circuit shown below, the capacitor is uncharged at time t = 0. If the switch is then closed, find the currents i₁ and i₂, and the charge on C at time t greater than zero.

Answer

NOTE: We could either:

• Set up the equations, take Laplace of each, then solve simultaneously
• Set up the equations, solve simultaneously, then take Laplace.

It is easier in this example to do the second method. In many examples, it is easier to do the first method.
For the first loop, we have:


For the second loop, we have:


Substituting (2) into (1) gives:


Next we take the Laplace Transform of both sides.
Note:

In this example, . So


Now taking Inverse Laplace:

And using result (2) from above, we have:

For charge on the capacitor, we first need voltage across the capacitor:

So, since , we have:

Graph of q(t):

EXAMPLE 4

In the circuit shown, the capacitor has an initial charge of 1 mC and the switch is in position 1 long enough to establish the steady state. The switch is moved from position 1 to 2 at t = 0. Obtain the transient current i(t) for t > 0.

Answer

Position 1, after a `long time': A
Position 2: ()
We apply emf, and consider the sum of the potential difference across elements.
In position 2, there is no emf.


Finding Laplace Transform:

Multiplying by :

Solving for I and completing the square on the denominator gives us:

So the transient current is:

We could transform the trigonometric part of this to a single expression:
2 cos 222.2t − 0.45 sin 222.2t = R cos(222.2t + α)


So

EXAMPLE 5

The system is quiescent. Find the loop current i₂(t).

Answer

Quiescent implies i₁, i₂ and their derivatives are zero for t = 0, ie
i₁(0) = i₂(0) = i₁'(0) = i₂'(0) = 0.

For loop 1:

For loop 2:


Substituting our result from (1) gives:

Taking Laplace transform:


Let
So


So
Taking Inverse Laplace:
So

Alternative answer using Scientific Notebook. (.tex file)

EXAMPLE 6

Consider a series RLC circuit where R = 20 W, L = 0.05 H and C = 10^-4 F and is driven by an alternating emf given by E = 100 cos 200t. Given that both the circuit current i and the capacitor charge q are zero at time t = 0, find an expression for i(t) in the region t > 0.

Answer

We use the following:

and obtain:

After multiplying throughout by 20, we have:

Taking Laplace transform and using the fact that i(0) = 0:

Using Scientific Notebook to find the partial fractions:


So

So
+ cos200t − 2 sin 200t

NOTE: Scientific Notebook can do all this for us very easily. In one step, we have:

+ cos200t − 2 sin 200t


Transient part:

Steady state part:

Alternative answer using Scientific Notebook. (.tex file)

EXAMPLE 7

A rectangular pulse v_R(t) is applied to the RC circuit shown. Find the response, v(t).

Graph of vR(t):

Note: v(t) = 0 V for all t < 0 s implies v(0^-) = 0 V.

Answer

Now
To solve this, we need to work in voltages, not current.
We start with .
The voltage across a capacitor is given by .
It follows that .
So for this example we have:

Substituting known values:

Then

Taking Laplace:
Since , we have:



So, taking inverse Laplace

NOTE: For the part: , we use:

So we have:

---

Solution Using Scientific Notebook

1. To find the Inverse Laplace:


2. To solve the original DE:

Exact solution for v(t):

To see what this means, we could write it as follows:

---

To see what our expression for v(t) means, we graph it as follows:

Alternative answer using Scientific Notebook. (.tex file)

http://www.intmath.com/laplace-transformation/10-applications.php