In the following circuit shown in figure, three diodes are connected each with forward resistance 40 and infinite backward resistance. The current through 100 resistance is

Step-by-step solution

The forward-biased diodes present identical 40 resistances, each combined in series with their respective resistors: 50 , 60 , and 160 , yielding branch resistances of 90 , 100 , and 200 . The equivalent resistance of their parallel combination is: 1 R_p = 1 90 + 1 100 + 1 200 = 47 1800 Thus, R_p = 1800 47 . The total resistance is: R_ total = 100 + 1800 47 = 6500 47 Applying Ohm’s law, the current through the 100 resistor is: I = 6 6500 47 = 282 6500 A 43.38 mA This value is closest to 36 mA among the provided opt