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A current of 6 A enters one corner P of ...

A current of 6 A enters one corner P of an equilateral triangle PQR having 3 wires of resistances 2 Q each and leaves by the corner R. Then the currents `I_(1) and I_(2)` are

A

2A, 4A

B

4A, 2A

C

1A , 2A

D

2A, 3A

Text Solution

Verified by Experts

The correct Answer is:
A
`I_(1)` will flow through `4 Omega and I_(2)` will flow through `2 Omega`. Since these are in parallel, `I_(1)=((2)/(2+4))xx6=2 A and I_(2)=4A`.
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