Question

4
2
78. A current of 6 A enters one corner P of an equilateral
triangle PQR having 3 wires
of resistances 2 2 each and
leaves by the corner R. Then the
currents I, and I, are
(a) 2 A, 4 A
(b) 4 A, 2 A
(c) 1A, 2A
(d) 2 A, 3 A​

Answer 1

I1 = 2 A

I2 = 4 A

Explanation:

Consider the figure given below  

                 

It is known that current always follows the least resistance path. As given in the question, current enters from point P and leaves from point R. I1 has to face double resistance as compare to I2 while reaching from P to R. Hence, I2 is two times than I1.