A current of 6 A enters one corner P of an e | vedclass

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(D) The current $I = 6 \ A$ enters at point $P$ and leaves at point $R$. At point $P$,the current splits into two paths: Path $1$: Through the branch

Vedclass · Accepted Answer

$2 \ A, 4 \ A$

Vedclass · Answer

(D) The current $I = 6 \ A$ enters at point $P$ and leaves at point $R$. At point $P$,the current splits into two paths: Path $1$: Through the branch $PQ$ and $QR$ in series. The resistance of this path is $R_1 = 2 \ \Omega + 2 \ \Omega = 4 \ \Omega$. Path $2$: Directly through the branch $PR$. The resistance of this path is $R_2 = 2 \ \Omega$. These two paths are in parallel between points $P$ and $R$. Using the current divider rule: $I_1 = I 	imes \left(\frac{R_2}{R_1 + R_2}ight) = 6 	imes \left(\frac{2}{4 + 2}ight) = 6 	imes \frac{2}{6} = 2 \ A$. $I_2 = I 	imes \left(\frac{R_1}{R_1 + R_2}ight) = 6 	imes \left(\frac{4}{4 + 2}ight) = 6 	imes \frac{4}{6} = 4 \ A$. Thus,$I_1 = 2 \ A$ and $I_2 = 4 \ A$.

$A$ current of $6 \ A$ enters one corner $P$ of an equilateral triangle $PQR$ having three wires of resistance $2 \ \Omega$ each and leaves by the corner $R$ as shown in the figure. Then the currents $I_1$ and $I_2$ are respectively:

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