enter image description here

  1. assume forward biased
  2. replace the diode with iD going down and short circuit it and make VD=0
  3. combine 70Ω and 30Ω resistors that are parallel and it becomes 21Ω
  4. now 21Ω resistance is in parallel with 3V and the (40Ω resistors and 50mA)
  5. ID=50mA meaning the assumption was correct

This is how i attempted the problem and i'm missing something because the final answer i get is odd. Any help or pointer would be greatly appreciated, Thankyou. update: work

update: enter image description here

  • \$\begingroup\$ Welcome to EE.SE! This appears to be a homework question. As such, you need to show us your work so far, and explain which part of the question you're having trouble with. For future reference: Homework questions on EE.SE enjoy/suffer a special treatment. We don't provide complete answers, we only provide hints or Socratic questions, and only when you have demonstrated sufficient effort of your own. Otherwise, we would be doing you a disservice, and getting swamped by homework questions at the same time. See also here. \$\endgroup\$ – Dave Tweed Sep 9 '18 at 2:23
  • \$\begingroup\$ It would be helpful in the future if you provided images that were already cleaned up, even if you have to re-write them. What you are presenting is a bit confusing with lines and parts scratched out. \$\endgroup\$ – user105652 Sep 9 '18 at 3:16
  • \$\begingroup\$ @Sparky256 i have updated \$\endgroup\$ – Rayin Sep 9 '18 at 5:10
  • \$\begingroup\$ check? @DaveTweed \$\endgroup\$ – Rayin Sep 10 '18 at 2:05
  • \$\begingroup\$ I have updated my work please verify? @DaveTweed \$\endgroup\$ – Rayin Sep 10 '18 at 4:07

It would probably be better to start by assuming the diode is open. This isolates the current source and the two series resistors, making it easy to check whether the anode of the diode is positive or negative with respect to the cathode.

Note that the 30Ω resistor is completely irrelevant to the rest of the circuit, since it is connected directly across a voltage source. You can ignore it altogether.

Note also that the power in the 40Ω resistor is determined entirely by the current source; nothing else in the circuit matters.

| improve this answer | |
  • \$\begingroup\$ did I do it correctly @Dave Tweed♦ ? \$\endgroup\$ – Rayin Sep 9 '18 at 2:36
  • \$\begingroup\$ No. If you assume that the diode is open, how did you get 0V on its anode? There's 50 mA flowing through those two resistors. \$\endgroup\$ – Dave Tweed Sep 9 '18 at 4:38
  • \$\begingroup\$ did I do it correctly @Dave Tweed♦? \$\endgroup\$ – Rayin Sep 9 '18 at 5:08
  • \$\begingroup\$ Yes, so far, so good. Now that you know that the diode is forward biased, what is the real value of \$V_A\$? What does this tell you about the reading on the voltmeter? (See the final hint in my answer above.) \$\endgroup\$ – Dave Tweed Sep 9 '18 at 12:03
  • \$\begingroup\$ Va is 3.09V but from point A prime to b prime... Vb is 3V while the point B prime is 0V so the voltmeter would be -3V...am I correct with my logic @Dave Tweed ♦ ? \$\endgroup\$ – Rayin Sep 10 '18 at 2:04

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