A current of 6 A enters one comer P of an equilateral triangle PQR having three wires of resistance $2…

A current of 6 A enters one comer P of an equilateral triangle PQR having three wires of resistance $2 \Omega$ each and leaves by the comer $R$ as shown in figure. Then the currents $\mathrm{I}_{\mathrm{I}}$, and $\mathrm{I}_2$, are respectively

1. $4 \mathrm{~A}, 2 \mathrm{~A}$
2. $3 \mathrm{~A}, 3 \mathrm{~A}$
3. $6 \mathrm{~A}, 0$
4. $2 \mathrm{~A}, 4 \mathrm{~A}$

Solution

The equivalent circuit diagram is
$\therefore \mathrm{I}_1=\left(\frac{2}{4+2}\right) \mathrm{I}=\frac{1}{3} \times 6=2 \mathrm{~A}$ $\mathrm{I}_2=\left(\frac{4}{4+2}\right) \mathrm{I}=\frac{2}{3} \times 6=4 \mathrm{~A}$

Asked in: AP EAMCET 2024 (20 May Shift 1)

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