A current of 6A enters one corner P of an equilateral triangle
Question:

A current of $6 \mathrm{~A}$ enters one corner $\mathrm{P}$ of an equilateral triangle $\mathrm{PQR}$ having 3 wires of resistance $2 \Omega$ each and leaves by the corner $R$. The currents $i_{1}$ in ampere is

Solution:

The current $\mathrm{i}_{1}=\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}\right)^{\mathrm{i}}$

$=\left(\frac{2}{4+2}\right) \times 6$

$\mathrm{i}_{1}=2 \mathrm{~A}$

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