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function ilamp=circuit872(RL)
M=[1 0 0 0 0 0;1 -1 0 -50 0 0;0 1 -1 0 -100 0;...
0 1 0 0 0 -300;0 0 1 0 -RL 0;0 0 0 1 -1 -1];
Vs=[5;0;0;0;0;0];
VI=M\Vs;
ilamp=VI(5);

Then, from the command window, we proceed by calling this function and
plotting the current in the lamp as a function of the resistance. Then we
graphically read for the value of R , which gives the desired current value.
L

In-Class Exercise

Pb. 8.5 For the circuit of Figure 8.1, find R that gives a 22-mA current in the
L
lamp. (Hint: Plot the current as function of the load resistor.)

8.7.3 ac Circuit Analysis
Conceptually, there is no difference between performing an ac steady-state
analysis of a circuit with purely resistive elements, as was done in Subsection
8.7.1, and performing the analysis for a circuit that includes capacitors and
inductors, if we adopt the tool of impedance introduced in Section 6.8, and
we write the circuit equations instead with phasors. The only modification
from an all-resistors circuit is that matrices now have complex numbers as
elements, and the impedances have frequency dependence. For convenience,
we illustrate again the relationships of the voltage-current phasors across
resistors, inductors, and capacitors:

˜
˜
V = IR (8.17)
R
˜
V = ˜ ( I j Lω ) (8.18)
L

˜
V = I ˜ (8.19)
C ω
(jC )
and restate Kirchoff’s laws again:

• Kirchoff’s voltage law: The sum of all voltage drops around a
closed loop is balanced by the sum of all voltage sources around
the same loop.

© 2001 by CRC Press LLC