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A current of 6 A enters one corner P of an equilateral triangle PQR having 3 wires of resistance $2;\Omega$ each and leaves by the corner R. The currents $i_1$ in ampere is:
The current divides: $R_1 = 4;\Omega$ (PQ + QR path) and $R_2 = 2;\Omega$ (PR path). $i_1 = \frac{R_2}{R_1 + R_2} \times 6 = \frac{2}{4+2} \times 6 = 2$ A.
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