can someone help me out?
http://maxpapers.com/wp-content/uploads/2012/11/9702_s14_ms_42.pdf ( question 10 biii)
http://onlineexamhelp.com/wp-content/uploads/2014/02/9702_w13_ms_43.pdf ( question 9 biii)
http://maxpapers.com/wp-content/uploads/2012/11/9702_s13_ms_42.pdf ( question 9b)
http://maxpapers.com/wp-content/uploads/2012/11/9702_w12_ms_43.pdf (question 9cii and 9dii)
For the first one
Vin = Votage at the non inverting input - voltage at the inverting input
gain=Vout/Vin
open loop gain is usually very large almost around (x 10^5) if not not specified
Votage at the non inverting input is constant at 1.0V (given) and the voltage at the inverting input is varying.
for the first few seconds voltage at the inverting input is more than that at non inverting input, therefore Vin is neagtive.
Vout will be a large neagtive number which is greater than the supply voltage hence op-amp will undergo negative saturation giving max Vout -5v and therfore diode which is reverse biased (R) will emit light
vice versa