Two coils of self-inductances 6 mH and 8 mH a | vedclass

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(C) Given, $L_{1} = 6 mH = 6 	imes 10^{-3} H$ and $L_{2} = 8 mH = 8 	imes 10^{-3} H$.For two coils connected in series with the highest coefficient

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$28$

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(C) Given, $L_{1} = 6 mH = 6 	imes 10^{-3} H$ and $L_{2} = 8 mH = 8 	imes 10^{-3} H$.For two coils connected in series with the highest coefficient of coupling $(k = 1)$, the equivalent self-inductance $L$ is given by the formula:$L = L_{1} + L_{2} + 2M$where $M$ is the mutual inductance, $M = k \sqrt{L_{1} L_{2}}$.For the highest coupling, $k = 1$, so $M = \sqrt{L_{1} L_{2}}$.Substituting the values:$L = L_{1} + L_{2} + 2 \sqrt{L_{1} L_{2}} 	ext{ (in } mH)$$L = 6 + 8 + 2 \sqrt{6 	imes 8}$$L = 14 + 2 \sqrt{48}$$L = 14 + 2 	imes 6.928$$L = 14 + 13.856 = 27.856 mH$Rounding to the nearest integer, we get $L \approx 28 mH$.

Two coils of self-inductances $6 mH$ and $8 mH$ are connected in series and are adjusted for the highest coefficient of coupling. The equivalent self-inductance $L$ for the assembly is approximately: (in $mH$)

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