Are you just throwing the problem out there expecting someone to supply you with an answer so that you can move on to the next problem?Here is the problem.
Need solution
Well i've solved but my answer is 0.1818 amps. your one seems differentAre you just throwing the problem out there expecting someone to supply you with an answer so that you can move on to the next problem?
If so, then here's an answer: 31.4 mA
If, instead, you want to learn how to solve the problem yourself, consider the following:
When they tell you to assume that the diode is made of silicon, what conclusions do you draw from that assumption?
What is the basic approach you have been taught to use when analyzing problems involving diodes?
Note that I only provided AN answer that you could use if all you wanted was someone to give you an answer.Well i've solved but my answer is 0.1818 amps. your one seems different
It depends on your assumption for the forward voltage drop of the diode. Every diode that you ever see, besides all the ones that you won't, will have a slightly different value for the forward voltage drop. Can you show your work?Well i've solved but my answer is 0.1818 amps. your one seems different
The question is also pretty clearly aimed at the very early days of a circuits course in which virtually everything has highly idealized behaviors so that the focus is on the analysis techniques.A silicon diode has a range of voltages. At 200mA a 1N4001 has a forward voltage of 0.8V and a 1N4148 has a little more than 1V.
The question probably assumes 0.7V which is wrong at the high current.
You did not allow for any current to go through the diode. KCL says some must go through the resistor and some must go through the diode. If V(A,B) is over 900 mV that diode is going to be on HARD!
You did not allow for any current to go through the diode.
This is probably not the answer being sought. The choice of component values makes it so that there are a couple of answers that result from using the wrong approach that are pretty close to the answer you get from using the correct approach. Need to see your work to know which category yours falls into.Well i've solved but my answer is 0.1818 amps. your one seems different
Maybe your view is accurate, but maybe you are assuming facts not in evidence. There is evidence that people abandon this site when they meet too much resistance. We'll see what happens.@Papabravo: Please avoid providing step by step solutions to a student's homework. It almost completely defeats the purpose. They have almost certainly seen similar problems worked by someone else in the text, in examples, and in lecture. Seeing yet one more problem worked by someone else is unlikely to suddenly fill in the gaps that exist. That will probably only happen as they struggle to work through a problem on their own with only hints and suggestions as to how to proceed past the current sticking point. What will likely happen by feeding them everything is that they will think they understand it, they will get credit for work that isn't their own, and then they will get reamed on the exam when they discover that seeing yet another problem worked by someone else didn't really address the gaps in their understanding.
Maybe you are making an assumption as well. Perhaps those abandoning the site wouldn’t find it of benefit in any case. Other than getting an answer without understanding and subsequently failing.Maybe your view is accurate, but maybe you are assuming facts not in evidence. There is evidence that people abandon this site when they meet too much resistance. We'll see what happens.
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