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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
Applying Kirchhoff's first law at the juction P, we get `6=I_(1)+I_(2)…(i)`
Applying Kirchhorff's second law to the closed loop PQRP, we get `-2I-2I_(1)+2I_(2)=0, or 4I_(1)-2I_(2)=O..(ii)`
Solve (i) and (ii), we get `I_(1)=2A, I_(2)=4A`
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