unable to solve this circuit......new to electronic

 

Hello "Help!!!"

Welcome to AAC!
Certainly you can come up with a better title. AAC receives dozens of requests for help daily. Imagine 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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base on my calculation,ID2 equal to 0.004A but not sure whether its correct...

 

Hello "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".
For "Homework Help" you need to show us your best effort before we can guide you with your homework.

thank you, will be extra careful next time

 

base on my calculation,ID2 equal to 0.004A but not sure whether its correct...

You need to show your work. How did you arrive at your answer?

 

You need to show your work. How did you arrive at your answer.

Hi,
9v-D1(0.7V)-D2(0.3V)=8V/2000ohm=0.004A

 

Hi,
9v-D1(0.7V)-D2(0.3V)=8V/2000ohm=0.004A

This is the current in D1. The current in D2 is less than the current in D1.

Your equation doesn't make sense. You have a voltage equated to a current.

This is what you meant:
\(\large I_{D1}=\frac{V_{SUP}-V_{D1}-V_{D2}}{R_1+R_3}=\frac{9V-0.7V-0.3V}{2k\Omega}=4mA\)

 

hi HC,
If there is 0.3V across R2, what is the current through R2.?
E

 

thank you guys, my latest answer is 0.004A.... after deduction of sum 4kohm

 

thank you guys, my latest answer is 0.004A.... after deduction of sum 4kohm

The question asks for the current in D2, not D1.

 

hi HC,
Are you sure it is precisely 4mA.?
E
Ref: post #8

 

9v-D1(0.7V)-D2(0.3V)=8V/4000ohm=0.002A

 

9v-D1(0.7V)-D2(0.3V)=8V/4000ohm=0.002A

How did you arrive at 4000ohm?

 

r1+r2+r3=4000ohm isn't it

How did you arrive at 4000ohm?

 

r1+r2+r3=4000ohm isn't it

That is incorrect.
You are just guessing and not using correct circuit theory.

 

let me try again

 

Post #6 and #7 are correct.
Now read post #8.

 

9v-D1(0.7V)-D2(0.3V)=8V/2000ohm=0.004A

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.

If you are studying physics you'd recognise that the voltage drop across a germanium diode is typically between 0.25V and 0.30V as it has a softer more gradual knee than the silicon diode which has a typical voltage drop of between 0.6 to 0.7V but it can be higher, and both depend on temperature. For a full answer you'd need the full voltage current characteristic curves of the diodes being used and (I think) an iterative approach to the final accurate answer.

Usually, 0.004A would be written as 4mA