# Solve the ideal diode circuit

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:

update:

• 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. – Dave Tweed Sep 9 '18 at 2:23
• 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. – user105652 Sep 9 '18 at 3:16
• @Sparky256 i have updated – Rayin Sep 9 '18 at 5:10
• check? @DaveTweed – Rayin Sep 10 '18 at 2:05
• I have updated my work please verify? @DaveTweed – Rayin Sep 10 '18 at 4:07

• 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.) – Dave Tweed Sep 9 '18 at 12:03