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Please see attached file. I just need a little help with understanding the concept.

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Answered Same Day Sep 22, 2022

Solution

Baljit answered on Sep 22 2022
50 Votes
ELECTRICAL ENGINEERING
Suppose cu
ent flowing through loop 1 is I1 and cu
ent flowing through loop I2. And VD6 is turn on voltage of D6 and VD7 is turn voltage for D7.
Apply Kirchhoff’s Voltage Law in Loop 1
V1- I1 R9-VD6 –(I1+I2)R11=0
KVL Law on Loop 2
V2-I2R10-VD7–(I1+I2)R11=0
1. ON/OFF conditions For Diodes:-
Given V1 and V2 are positive
so
V1- I1 R9- VD6 -(I1+I2)R11=0
V2-I2R10-VD7 -(I1+I2)R11=0
Now Suppose Voltage drop across R11 is VR11 which is equal to (I1+I2)R11.
Case 1:-Both diodes are turn ON means in forward bias if
V1- I1 R9- VD6> VR11
V2-I2R10-VD7...
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