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    An RL Circuit with Switches at T=0 Switch S1 Opens

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    An RL circuit with switches is shown in the figure. Switch s1 remains closed and s2 remains open for a long time (steady-state). At t=0 switch s1 opens and s2 closes. Find V_1(t) for t > 0.

    © BrainMass Inc. brainmass.com December 24, 2021, 4:58 pm ad1c9bdddf
    https://brainmass.com/engineering/electronic-engineering/circuit-switches-switch-opens-19655

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    SOLUTION This solution is FREE courtesy of BrainMass!

    Answer is B. Because

    During steady state, S2 is open, means; the last inductor is cut off from the circuit. So, what we have is, 6 Ohm, 4 ohm and 1 H. And since, it's the DC current which is flowing, inductor acts a short circuit at steady state.

    Hence we have 5A current source in parallel with 6 and 4 ohm resistors which are in parallel with each other. So, the voltage developed at node V1, when V1 is close is,

    5A * [(6*4)/(6+4)] = 12 V.

    But, at t=0, V1 opens and V2 closes. So we have inductor in the circuit. But the 6 ohm resistor and the 5A source are cut off from the circuits. Effectively we have an RL tank circuit. with time constant T = 1/[R/L] =1/[4/1] = 1/4

    Where,

    L= 1H ; since we are asked to calculate V2 at t=0. So only the first inductor comes into picture and the other inductor is shorted at that point of time.

    So, the voltage V2, at t=0 is

    -12 exp (-t/T) = -12 exp (-4t)

    -Ve sign is due to the current flowing out of the node V2.

    B is your ans.

    Hope this helped.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 4:58 pm ad1c9bdddf>
    https://brainmass.com/engineering/electronic-engineering/circuit-switches-switch-opens-19655

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