Physics-
General
Easy

Question

A current of A enters one corner one corner P of an equilateral triangle P Q Rhaving 3 wires of resistance 2blank capital omegaeach and leaves by the corner R. then the current I subscript 1 end subscriptand I subscript 2 end subscriptare

  1. 2A, 4A    
  2. 4A, 2A    
  3. 1A, 2A    
  4. 2A, 3A    

The correct answer is: 2A, 4A


    From Kirchhoff’s first law at junction P,
    I subscript 1 end subscript plus I subscript 2 end subscript equals 6 blank horizontal ellipsis open parentheses i close parentheses
    From Kirchhoff’s second law to the closed circuit PQRP,
    negative 2 I subscript 1 end subscript minus 2 I subscript 1 end subscript plus 2 I subscript 2 end subscript equals 0
    ⟹ negative 4 I subscript 1 end subscript plus 2 I subscript 2 end subscript equals 0
    ⟹ 2 I subscript 1 end subscript minus I subscript 2 end subscript equals 0 horizontal ellipsis open parentheses i i close parentheses
    Adding Eqs. (i) and (ii), we get
    3 I subscript 1 end subscript equals 6 ⟹ I subscript 1 end subscript equals 2 A
    From Eq. (i),
    I subscript 2 end subscript equals 6 minus 2 equals 4 A

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