A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 2 years ago
A silicon diod is in series with a 1 kilo ohm resistor and a 5 volt battery.If the anode is connected to the positive battery terminal the cathode voltage with respect to the negative battery terminal is:
A) 0.7v
B) 0.3v
C) 5.7 v
D) 4.3v
anonymous
 2 years ago
A silicon diod is in series with a 1 kilo ohm resistor and a 5 volt battery.If the anode is connected to the positive battery terminal the cathode voltage with respect to the negative battery terminal is: A) 0.7v B) 0.3v C) 5.7 v D) 4.3v

This Question is Closed

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.04.3V. The forward voltage drop across standard silicon diodes is 0.7V (same as a BJT's baseemitter drop, incidentally). 5V from the source minus 0.7V from the diode leaves 4.3V across the resistor. Incidentally, you can now use ohm's law on the resistor to determine the current through it. Since the current through series elements will be equal, you also now know the current through the diode. This is very useful in sizing resistors for LED circuits.

KenLJW
 2 years ago
Best ResponseYou've already chosen the best response.0The diode voltage depends on the current through it and the diode design. Generally it will be less the .9 V and greater then 0 when forward biased. So the answer to this question could be a diode drop of .3 or .7 dependent on diode design and the voltage drop across the 1K resistor.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.