 Source-fed parallel resistor-inductor switching

ID: source-fed-parallel-resistor-inductor-switching

V_{0}

R_{1}

feed parallel branches

L

and

R_{2}

R_{th} = R_{1} R_{2} / (R_{1} + R_{2})

and

V_{th} = V_{0} R_{2} / (R_{1} + R_{2})

. Starting with zero inductor current,

i_{L} = (V_{0} / R_{1}) (1 - e^{- t / τ})

and

i_{2} = V_{0} e^{- t / τ} / (R_{1} + R_{2})

R_{2}

L V_{0}^{2} / [2 R_{1} (R_{1} + R_{2})]

. Disconnecting the source leaves the stored magnetic energy

L i_{L}^{2} /2

to dissipate through

R_{2}

L / R_{2}

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