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Two coils are mutually coupled, with L1=50 mH, L2=120 mH, and k=0.5. Calculate the maximum possible equivalent inductance if the coils are connected in parallel

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Answered 1 year ago
Answered 1 year ago

b) There are also two possibilities.

  1. The polarity of the mutually-induced voltage is positive and in that case

Leq=L1L2M2L1+L22M like we show in the previous problem (13.4) .\begin{align*} L_{eq} &=\frac{L_1L_2-M^2}{L_1+L_2-2M} \ \text{like we show in the previous problem (13.4) .} \end{align*}

  1. The polarity of the mutually-induced voltage is negative and in that case

Leq=L1L2(M)2L1+L22(M)Leq=L1L2M2L1+L2+2M\begin{align*} L_{eq} &=\frac{L_1L_2-(-M)^2}{L_1+L_2-2(-M)} \\ L_{eq} &=\frac{L_1L_2-M^2}{L_1+L_2+2M} \end{align*}

LeqL_{eq} is greater in first case

L1+L22M<L1+L2+2M L1L2M2L1+L22M>L1L2M2L1+L22M\begin{align*} L_1+L_2-2M<L_1+L_2+2M \\ \Rightarrow \ \frac{L_1L_2-M^2}{L_1+L_2-2M} >\frac{L_1L_2-M^2}{L_1+L_2-2M} \\ \end{align*}

So

Leq=48.63 mH\begin{align*} L_{eq}=\boxed{48.63\mathrm{~mH}} \end{align*}

The circuit with the maximum possible equivalent inductance is shown below.

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