Three coils of inductance L_1 = 2 H, L_2 = 3 | vedclass

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(D) For inductors connected in series, the effective inductance is $L_{eff} = L_1 + L_2 + L_3$. For inductors connected in parallel, the effective ind

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$Q$

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(D) For inductors connected in series, the effective inductance is $L_{eff} = L_1 + L_2 + L_3$. For inductors connected in parallel, the effective inductance is given by $\frac{1}{L_{eff}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}$.$1$. Combination $P$: The inductors are in series. $L_{eff} = 2 + 3 + 6 = 11 \ H$.$2$. Combination $Q$: The inductors are in parallel. $\frac{1}{L_{eff}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = \frac{6}{6} = 1 \ H^{-1}$. Thus, $L_{eff} = 1 \ H$.$3$. Combination $R$: $L_1$ and $L_2$ are in series, and this combination is in parallel with $L_3$. $L_{series} = 2 + 3 = 5 \ H$. Then $\frac{1}{L_{eff}} = \frac{1}{5} + \frac{1}{6} = \frac{11}{30}$, so $L_{eff} = \frac{30}{11} \approx 2.72 \ H$.$4$. Combination $S$: This is a mixed series-parallel circuit. $L_1$ is in parallel with $L_3$, and this combination is in series with $L_2$. $\frac{1}{L_{parallel}} = \frac{1}{2} + \frac{1}{6} = \frac{3+1}{6} = \frac{4}{6} = \frac{2}{3} \ H^{-1}$, so $L_{parallel} = 1.5 \ H$. Then $L_{eff} = 1.5 + 3 = 4.5 \ H$.Therefore, the correct combination is $Q$.

Three coils of inductance $L_1 = 2 \ H$, $L_2 = 3 \ H$, and $L_3 = 6 \ H$ are connected such that they are separated from each other. To obtain an effective inductance of $1 \ H$, which of the following combinations, as shown in the figure, is correct?

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