VD А. Given the following circuit in Figure ww 5 ΚΩ determine whether the Diode is conducting or not. You need to assume both cases. Firstly, assume that the Diode is conducting and do the needed calculations, then assume that it is not conducting and do the A 10 K 2 needed calculations. Also, you need to show your drawing on how the circuit is changed in 1000 N both cases. 20 V Vin The first assumption with needed calculations 15 V The second assumption with needed calculations ( The drawing in both cases (

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VD
А.
Given the following circuit in Figure
ww
5 ΚΩ
determine whether the Diode is
conducting or not. You need to assume both cases. Firstly, assume that the Diode is
conducting and do the needed calculations, then assume that it is not conducting and do the
A
10 K 2
needed calculations. Also, you need to show your drawing on how the circuit is changed in
1000 N
both cases.
20 V
Vin
The first assumption with needed calculations
15 V
The second assumption with needed calculations (
The drawing in both cases (
Transcribed Image Text:VD А. Given the following circuit in Figure ww 5 ΚΩ determine whether the Diode is conducting or not. You need to assume both cases. Firstly, assume that the Diode is conducting and do the needed calculations, then assume that it is not conducting and do the A 10 K 2 needed calculations. Also, you need to show your drawing on how the circuit is changed in 1000 N both cases. 20 V Vin The first assumption with needed calculations 15 V The second assumption with needed calculations ( The drawing in both cases (
Expert Solution
Step 1

Given data

The given circuit

R1=5 R2=10 R3=1000 Ω=1 V1=20 V V2=15 V

 

Step 2

Explanation:

The first assumption is diode conducting.

When diode conducts,  then circuit is short and diode is considered to be ideal.

Draw the circuit diagram

Electrical Engineering homework question answer, step 2, image 1

 

The diode conduct the voltage across vD=0.

Here, R1 and R2 are in the parallel, equivalent resistance is,

r=R1×R2R1+R2

Substitute the given values in the above expression.

r=5×105+10=5015=103 

The expression for the total current is,

I=20-15r+R3

Substitute the above values.

I=5103+1=1513   1.154 mA

Using current division rule

ID=I×R2R1+R2ID=1.154×1010+5    =0.769 mA

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