Symbolic/Examples/Circuits

Brute Force 1

The following example shows the "brute force" method of setting up and solving for all the element currents and voltages for a simple resistive circuit and then using those solutions to solve for auxiliary information (in this case, some powers). The circuit involved is:

Equations

Element Equations

Mainly, these are Ohm's Law equations for the resistors, so:

\( \begin{align} R_1:& & v_1&=i_1R_1\\ R_2:& & v_2&=i_2R_2 \end{align} \)

KCL Equations

The number of independent KCL equations s one less than the number of nodes, so in this case, 2. Note: all three nodal KCL equations are written below, but node \(n_c\)'s is not used in the Maple worksheet.

\( \begin{align} KCL,n_a:& & -i_a+i_1+i_2&=0\\ KCL,n_b:& & -i_2+i_b&=0\\ KCL,n_c:& & i_a-i_1-i_b&=0 \end{align} \)

KVL Equations

The number of independent KVL equations is equal to the number of meshes for a 2-D circuit, or to the number of elements, minus the number of nodes, plus one for circuits in general. In this case, that is 2 independent KVL (two meshes, or 3 elements - 3 nodes + 1 = 2). For the brute force method, just use the mesh equations:

\( \begin{align} KVL,l_1:& & -v_a+v_1&=0\\ KVL,l_2:& & -v_1+v_2+v_b&=0 \end{align} \)

Auxiliary Equations

For this example, the auxiliary equations will be used to determine the power delivered by each source and the power absorbed by each resistor:

\( \begin{align} p_{del,i_a}&=v_ai_a & p_{del,v_b}&=-v_bi_b \\ p_{abs,R_1}&=v_1i_1 & p_{abs,R_2}&=v_2i_2 \end{align} \)

Note that all elements except for \(i_a\) are labeled passively.

Code

The code for this example is available for download: Circuit1Demo.mw. You can also look at a PDF of the code. The code assumes that:

\( \begin{align} R_1&=1000~\Omega & R_2&=2200~\Omega\\ i_a&=0.005~\mbox{A} & v_b&=12~\mbox{V} \end{align} \)