Question

4.64 pspice multisim find the norton equivalent with respect to the terminals a, b for the circuit in fig. p4.64.
figure p4.64

Explanation:

Step1: Calculate the Norton current ($I_N$)

First, short - circuit terminals a and b. The current through the short - circuit is the Norton current. The 15 A current source divides between the two parallel paths. The equivalent resistance of the two parallel paths (one with $2\Omega$ and $12\Omega$ in series, and the other with $8\Omega$ and $10\Omega$ in series) is considered. The resistance of the first path $R_1=2 + 12=14\Omega$, and the resistance of the second path $R_2=8 + 10 = 18\Omega$. Using the current - division rule, $I_N=15\times\frac{R_2}{R_1 + R_2}=15\times\frac{18}{14 + 18}=15\times\frac{18}{32}=\frac{135}{16}=8.4375$ A.

Step2: Calculate the Norton resistance ($R_N$)

Open - circuit the current source. Then, find the equivalent resistance seen from terminals a and b. The $2\Omega$ and $8\Omega$ resistors are in parallel, and the $12\Omega$ and $10\Omega$ resistors are in parallel. The equivalent resistance of $2\Omega$ and $8\Omega$ in parallel is $R_{28}=\frac{2\times8}{2 + 8}=1.6\Omega$. The equivalent resistance of $12\Omega$ and $10\Omega$ in parallel is $R_{1210}=\frac{12\times10}{12 + 10}=\frac{120}{22}=\frac{60}{11}\Omega$. Then $R_N=R_{28}+R_{1210}=1.6+\frac{60}{11}=\frac{17.6+60}{11}=\frac{77.6}{11}\approx7.0545\Omega$.

Answer:

The Norton current $I_N = 8.4375$ A and the Norton resistance $R_N\approx7.0545\Omega$