I = 2.99e-002 A
For this current,V = 0.2 V
(a)Therefore I = 29.8 mA
(b)If battery is inserted with reverse polarity,voltage drop across the 1.5 K resistors is only 15 mV and may be neglected
(c)In forward direction, I=29.8 mA
In reverse direction we draw a load line from V=-30 V to I=-30 mA
Then,I = -20 mA
Current=20 mA as there is a 10 V drop