A current of 6, A enters at corner P of an eq | vedclass

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(A) The circuit consists of two parallel branches connected between points $P$ and $R$. Branch $1$ consists of the wire $PQ$ and $QR$ in series. The r

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$2$

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(A) The circuit consists of two parallel branches connected between points $P$ and $R$. Branch $1$ consists of the wire $PQ$ and $QR$ in series. The resistance of this branch is $R_1 = 2 \,\Omega + 2 \,\Omega = 4 \,\Omega$. Branch $2$ consists of the wire $PR$ directly,with resistance $R_2 = 2 \,\Omega$. The total current $I = 6 \, A$ divides into these two parallel branches. Using the current divider rule,the current $i_1$ flowing through the branch $PQ$ (and then $QR$) is given by: $i_1 = I 	imes \frac{R_2}{R_1 + R_2}$ $i_1 = 6 	imes \frac{2}{4 + 2} = 6 	imes \frac{2}{6} = 2 \, A$. Thus,the current $i_1$ is $2 \, A$.

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

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