ECON 211 Problems Chapter 10, Fall 2008
American University of Beirut · Elementary Microeconomics Theory · 6 pages
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American University of Beirut
Electrical and computer engineering department
EECE 312
Problem session 4
3/11/2008
Chapter 10 problems 7, 21, 23, 37, 48
Solution:
first method, you can solve the problem without thevenin: write KCL: ix = ix/2 +i ==>
i=ix/2, then write KVL: 5-ix*1kohm -v=0 ==> 5 - 2*i -v=0 This is the load line that you
can draw on the diode charcteristic for graphical analysis.
second method: to find thevenin eq. circuit:
open the diode and find v across the o.c. (= Vth). Since in this case, you will have ix and
ix/2 flowing in the same loop, then ix =0, and v = Vth=5V.
to get Rth, zero the voltage source, and apply at the place of the diode a Vtest, and call
the current going out of it Itest. Rth = Vtest/Itest.
in this circuit, Vtest = -ix*1kohm, and itest + ix = ix/2 (KCL) therefore
itest = -ix/2. replace in Vtest : Vtest = 2*itest therefore Rth = Vtest/Itest=2kohm.