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By the way, D1 is reverse bias, sorry about the bad drawing
If D1 is reversed biased we can remove it safely Now our equivalent diagram looks like following|dw:1335316744178:dw| Now apply Kirchoof's law in the loops ABEFE & BCDEB We have three unknowns I, I1 & ID2 The third equations comes from the junction E where I=I1+ID2
You did not provide the value of the battery voltage so it is difficult to calculate the exact current in this circuit. The current thru D2 will depend on this. Also the voltage across D1 which is the IR drop across the 1.3 K resistor. The current thru the 3K plus the 1.3K (4.4K) would be the battery voltage minus the 0.7 volt across D2. divided by the resistance. The current across the 20K resistor is limited by the diode D2
The current thru the 3k and 1.3k is: (5V-.7V)/(3K + 1.3K)= 1 milliamp thus the voltage across the reverse biased D1 would then be 1.3 Volt. The current through the 20K resistor is .7/20K Ohm=35 micro-amps and is not enough to be significant.
Technically the 35 microamps would be added to the 1 milliamp and the voltage across would be up and additional 45 millivolts
Thanks for your help Radar and RS!
i love open study
It does prove to be helpful. I learn quite a bit from the posts and answers