A current of 9text{A} enters point P of an eq | vedclass

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(B) The total current of $9	ext{A}$ enters at point $P$ and splits into two paths: one path goes directly through the branch $PR$ (carrying current $

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$3	ext{A}, 6	ext{A}$

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(B) The total current of $9	ext{A}$ enters at point $P$ and splits into two paths: one path goes directly through the branch $PR$ (carrying current $I_2$), and the other path goes through the branch $PQ$ and then $QR$ to reach $R$ (carrying current $I_1$).Since the triangle is equilateral and each side has a resistance of $3\Omega$, the resistance of the branch $PR$ is $R_1 = 3\Omega$.The resistance of the path $PQR$ is $R_2 = R_{PQ} + R_{QR} = 3\Omega + 3\Omega = 6\Omega$.According to the current divider rule, the current in a branch is inversely proportional to its resistance.$I_1 = I_{total} 	imes \frac{R_1}{R_1 + R_2} = 9 	imes \frac{3}{3 + 6} = 9 	imes \frac{3}{9} = 3	ext{A}$.$I_2 = I_{total} 	imes \frac{R_2}{R_1 + R_2} = 9 	imes \frac{6}{3 + 6} = 9 	imes \frac{6}{9} = 6	ext{A}$.Thus, $I_1 = 3	ext{A}$ and $I_2 = 6	ext{A}$.

$A$ current of $9\text{A}$ enters point $P$ of an equilateral triangle $PQR$ having three wires of $3\Omega$ each and leaves by point $R$. The currents $I_1$ and $I_2$ are respectively

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