Homework Given January 16, 2007

Due January23, 2007

These aren't homework problems per se; the Exercises are intended as a study aid for those reading the chapter.  These are selected to make sure that the intended points stick with the student.

Live Links to the Exercises

*      Exercise 1.1

*      Exercise 1.5

*      Exercise 1.12

Exercise 1.1

Plot the voltage-current curve of resistors of 1 k, 5 k, and 20 k,


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Exercise 1.5

Find the value of  in Figure 1.5 if





It's tempting to use a Norton equivalent of , , and , because that would result in two current sources in parallel across two resistors in parallel, a very simple circuit to solve.  However, the section is on superposition, so we will solve it by superposition.


The component of  from  is


and the component  from  is identical,


The component of  from  is


The final value of  is the sum of the components from , , and ,


For the values given in the problem statement,  is



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Exercise 1.12

Plot the v-I equation of the circuit of Figure 1.11 if the value of  is changed to


The figure is part of Example 1.3, which is a superposition example.  To follow the intent of the exercise, we will solve it by superposition.


Note that the zig-zag transistor model geometry of Figure 1.11 has been changed here to the more familiar ladder configuration that we use for circuit analysis.  To plot the v-I characteristic of  and , we need to find the open circuit voltage and short circuit current.  In class, I solved this circuit directly instead of using superposition as in Example 1.3, which is the context for the figure.  Here, we use superposition.


The component of the open circuit voltage  due to  is found by finding  and noting that the controlled current source and  form a Norton equivalent across the terminals which  is measured.  This we have





The short circuit current is simply the controlled current source current, negated because  is positive out the terminal, opposite the direction of the arrow in the controlled current source,



The voltage source  drives current through  opposite than that of  but its effect on the controlled current source is otherwise identical, so we have




By superposition, the open circuit voltage is the sum of the two voltages due to  and , and the short circuit current is the sum of the currents due to  and ,





For the values given in the problem,






and, finally, the total open circuit voltage and short circuit current are




We plot the v-i curve from the open circuit voltage, when current is zero, and the short circuit current, when the voltage is zero, as a straight line between these points as below.


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