I know it very simple to understand. What i have understood is as follows
Their would be a potential difference across R3 nd R4.
* thats correct, the Vs1 = 1V according to the drawing
When the +V is given to R1 and ground to R2,
*the R2 is not grounded, it goes to ground via R6 and R8
the diode D2 and D3 will conduct providing again a potential difference at the end of the ciruict which could be used for the difference amplifier.
*The drawing shows Vs1 is +1V.
* With the centre points of the diodes connected to 0V ground via 1k resistors R7 and R8 and to the 1V source via R1/R5 and R2/R6 which are 100k resistors, the voltage at the junction of the diodes will be less than the turn on voltage [0.7] of the diodes, so the diodes will not be conducting.
When the +V is given to R2 and ground to R1, D4 and D1 will conduct giving potential difference at the end.
The point i stuck is when the +V is given to R1, D2 also conduct. So what would be the voltage at that point? what would be the equation for that??
* with voltage drop across R1/R5 and R2/R6 the diodes are off and can be ignored, so just calculate the circuit using the resistor voltage drops.
hi,
Hope you follow ok.
IMO its a poorly designed question.