Equivalent circuits
Suppose we connect a generator $\emf$ to a circuit containing some complicated interconnection of impedances, as indicated schematically in Fig. 22–15(a). All of the equations we get from Kirchhoff’s rules are linear, so when we solve them for the current $I$ through the generator, we will get that $I$ is proportional to $\emf$. We can write \begin{equation*} I=\frac{\emf}{z_{\text{eff}}}, \end{equation*}  where now $z_{\text{eff}}$ is some complex number, an algebraic function of all the elements in the circuit. (If the circuit contains no generators other than the one shown, there is no additional term independent of $\emf$.) But this equation is just what we would write for the circuit of Fig. 22–15(b). So long as we are interested only in what happens to the left of the two terminals $a$ and $b$, the two circuits of Fig. 22–15 are equivalent. We can, therefore, make the general statement that any two-terminal network of passive elements can be replaced by a single impedance $z_{\text{eff}}$ without changing the currents and voltages in the rest of the circuit. This statement is of course, just a remark about what comes out of Kirchhoff’s rules—and ultimately from the linearity of Maxwell’s equations.  The idea can be generalized to a circuit that contains generators as well as impedances. Suppose we look at such a circuit “from the point of view” of one of the impedances, which we will call $z_n$, as in Fig. 22–16(a). If we were to solve the equation for the whole circuit, we would find that the voltage $V_n$ between the two terminals $a$ and $b$ is a linear function of $I_n$, which we can write \begin{equation} \label{Eq:II:22:22} V_n=A-BI_n, \end{equation}  where $A$ and $B$ depend on the generators and impedances in the circuit to the left of the terminals. For instance, for the circuit of Fig. 22–13, we find $V_1=I_1z_1$. This can be written [by rearranging Eq. (22.20)] as \begin{equation} \label{Eq:II:22:23} V_1=\biggl[ \biggl(\frac{z_2}{z_2+z_3}\biggr)\emf_2-\emf_1 \biggr]-\frac{z_2z_3}{z_2+z_3}\,I_1. \end{equation}  The complete solution is then obtained by combining this equation with the one for the impedance $z_1$, namely, $V_1=I_1z_1$, or in the general case, by combining Eq. (22.22) with \begin{equation*} V_n=I_nz_n. \end{equation*}  If now we consider that $z_n$ is attached to a simple series circuit of a generator and an impedance, as in Fig. 22–16(b), the equation corresponding to Eq. (22.22) is \begin{equation*} V_n=\emf_{\text{eff}}-I_nz_{\text{eff}}, \end{equation*}  which is identical to Eq. (22.22) provided we set $\emf_{\text{eff}}=A$ and $z_{\text{eff}}=B$. So if we are interested only in what happens to the right of the terminals $a$ and $b$, the arbitrary circuit of Fig. 22–16 can always be replaced by an equivalent combination of a generator in series with an impedance. 