The circuit shown in following figure contains two diode $D_1$ and $D_2$ each with a forward resistance of 50 ohms and with infinite backward resistance. If the battery voltage is 6 V, the current through the 100 ohm resistance (in amperes) is

0.02

According to the given polarity, diode $D_1$ is forward biased while $D_2$ is reverse biased. Hence current will pass through $D_1$ only.

So current $i=\frac{6}{(150+50+100)}=0.02A$