thank you, will be extra careful next timeHello "Help!!!"
Welcome to AAC!
Certainly you can come up with a better title. AAC receives dozens of requests for help daily. Imaging how useful a title of "Help!!!" is if all threads were titled "Help!!!".
Besides that, your post has been moved to "Homework Help".
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You need to show your work. How did you arrive at your answer?base on my calculation,ID2 equal to 0.004A but not sure whether its correct...
Hi,You need to show your work. How did you arrive at your answer.
This is the current in D1. The current in D2 is less than the current in D1.Hi,
9v-D1(0.7V)-D2(0.3V)=8V/2000ohm=0.004A
The question asks for the current in D2, not D1.thank you guys, my latest answer is 0.004A.... after deduction of sum 4kohm
How did you arrive at 4000ohm?9v-D1(0.7V)-D2(0.3V)=8V/4000ohm=0.002A
How did you arrive at 4000ohm?
That is incorrect.r1+r2+r3=4000ohm isn't it
Are you studying physics or engineering? If you are studying engineering you were nearly right, if you'd added that the current through R2 is small enough (an order of magnitude smaller) to ignore, especially as your voltage drops across the diodes are only roughly right.9v-D1(0.7V)-D2(0.3V)=8V/2000ohm=0.004A
by Jake Hertz
by Aaron Carman
by Jake Hertz