What is Linear dipole, linear circuit? — Physics

Definition 6.16 University Physics — Year 1 · Chapter 6 — DC Circuits: Kirchhoff’s Laws and Theorems

A dipole is linear if its characteristic is a straight line (resistor, ideal and real sources); a circuit made only of linear dipoles is a linear circuit. Its unknowns obey a system of linear equations (Kirchhoff’s laws plus the characteristics), whose solution is unique.

Two sources feeding a common load R_3. Seen from R_3’s terminals, the rest of the circuit is a Thévenin source (E_ Th, R_ Th); Millman’s theorem gives the potential of node N in one line. — Two sources feeding a common load $R_3$ . Seen from $R_3$ ’s terminals, the rest of the circuit is a Thévenin source ( $E_{\mathrm{Th}}$ , $R_{\mathrm{Th}}$ ); Millman’s theorem gives the potential of node $N$ in one line.

Examples

Example 6.20 (Three ways to the same answer)

In the figure, $E_1 = 10\,\mathrm{V}$ , $R_1 = 1.0\,\mathrm{k}\Omega$ , $E_2 = 5.0\,\mathrm{V}$ , $R_2 = 2.0\,\mathrm{k}\Omega$ , $R_3 = 2.0\,\mathrm{k}\Omega$ . Millman: $V_N = (10/1 + 5/2 + 0/2)/(1 + 0.5 + 0.5) = 12.5/2 = 6.25\,\mathrm{V}$ . Superposition: $E_1$ alone sees $R_2 \parallel R_3 = 1\,\mathrm{k}\Omega$ , so $u = 10 \times 1/2 = 5\,\mathrm{V}$ ; $E_2$ alone sees $R_1 \parallel R_3 = 0.667\,\mathrm{k}\Omega$ , $u = 5 \times 0.667/2.667 = 1.25\,\mathrm{V}$ ; total $6.25\,\mathrm{V}$ . Thévenin seen from $R_3$ : $E_{\mathrm{Th}}$ is the open-circuit voltage of the divider $(E_1, R_1, R_2, E_2)$ : $E_2 + (E_1 - E_2)R_2/(R_1 + R_2) = 5 + 5 \times 2/3 = 8.33\,\mathrm{V}$ ; $R_{\mathrm{Th}} = R_1 \parallel R_2 = 0.667\,\mathrm{k}\Omega$ ; then $u = 8.33 \times 2/2.667 = 6.25\,\mathrm{V}$ .

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