Question

A current of 6 A enters one corner P of an equilateral triangle PQR having 3 wires of resistances 2 $\sqrt{\sigma_{1}\sigma_{2}}$each and level by the corner R. Then the current I1 and I2 are

• A. 2A, 4A
• B. 4A, 2A
• C. 1A, 2A
• D. 2A, 3A

Answer 1

Answer: (A)

2A, 4A

Answer 2

: Applying Kirchhoff’s first law at the junction P, we get,

$6 = I_{1} + I_{2}$ …(i)

Applying Kirchhoff’s second law to the closed loop PQRP, we get

$- 2I_{1} - 2I_{1} + 2I_{2} = 0$ or $2I_{1} + 2I_{1} - 2I_{2} = 0$ or,

$4I_{1} - 2I_{2} = 0$ …(ii)

Solve (i) and (ii), we get,$I_{1} = 2A,I_{2} = 4A.$