ECD2601
PORTFOLIO
OCT/NOV EXAM
2026
UNIQUE NO:
DUE DATE: 5 Days
, ENGINEERING GRAPHICS FOR ELECTRICAL DESIGN
PROJECTS - ECD2601
OCTOBER/NOVEMBER 2026
SEMESTER COURSE PORTFOLIO EXAM
Question 1
Determine the voltage across a diode in Figure 1, using the Ideal model, practical
model, and complete model. Assume $r_d = 5\ \Omega$ and $r_R' = 10\text{
M}\Omega$.
Circuit Analysis & Diode State:
o Looking at Figure 1, we have a $10\text{ V}$ source on the left connected
via a $10\text{ k}\Omega$ resistor to the anode of the diode, and a
$20\text{ V}$ source on the right connected via a $10\text{ k}\Omega$
resistor to the cathode.
o The ground is at the bottom node, connecting the negative terminal of the
$10\text{ V}$ source, the diode cathode/anode reference, and the positive
terminal of the $20\text{ V}$ source. Wait, let's look closely at the battery
polarities in Figure 1:
Left battery: Top is positive ($+10\text{ V}$), bottom is negative.
Right battery: Top is negative, bottom is positive ($+20\text{ V}$ at
the bottom, so the top node connected to the resistor is at $-
20\text{ V}$ relative to ground).
o Let's determine the open-circuit voltages at the junction above the diode:
If the diode is considered open:
Left branch current $I_{\text{left}} = \frac{10\text{ V}}{10\text{
k}\Omega} = 1\text{ mA}$ flowing toward the node.
Right branch connected to the top of the right $10\text{
k}\Omega$ resistor: The right battery has its negative
PORTFOLIO
OCT/NOV EXAM
2026
UNIQUE NO:
DUE DATE: 5 Days
, ENGINEERING GRAPHICS FOR ELECTRICAL DESIGN
PROJECTS - ECD2601
OCTOBER/NOVEMBER 2026
SEMESTER COURSE PORTFOLIO EXAM
Question 1
Determine the voltage across a diode in Figure 1, using the Ideal model, practical
model, and complete model. Assume $r_d = 5\ \Omega$ and $r_R' = 10\text{
M}\Omega$.
Circuit Analysis & Diode State:
o Looking at Figure 1, we have a $10\text{ V}$ source on the left connected
via a $10\text{ k}\Omega$ resistor to the anode of the diode, and a
$20\text{ V}$ source on the right connected via a $10\text{ k}\Omega$
resistor to the cathode.
o The ground is at the bottom node, connecting the negative terminal of the
$10\text{ V}$ source, the diode cathode/anode reference, and the positive
terminal of the $20\text{ V}$ source. Wait, let's look closely at the battery
polarities in Figure 1:
Left battery: Top is positive ($+10\text{ V}$), bottom is negative.
Right battery: Top is negative, bottom is positive ($+20\text{ V}$ at
the bottom, so the top node connected to the resistor is at $-
20\text{ V}$ relative to ground).
o Let's determine the open-circuit voltages at the junction above the diode:
If the diode is considered open:
Left branch current $I_{\text{left}} = \frac{10\text{ V}}{10\text{
k}\Omega} = 1\text{ mA}$ flowing toward the node.
Right branch connected to the top of the right $10\text{
k}\Omega$ resistor: The right battery has its negative
