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If the value of R is 5 ohm in the circuit and the switch is closed for a long time and then the switch is opened at t = 0 then what will be the voltage across the inductor? Will it be equal to the battery voltage at t = 0?

Note that I just need intuition. It's not home work. I just want to know that what will be the voltage across the inductor at t = 0. The current across the inductor I know will be 4 ampere as it was before the switch is opened but I am confused about the voltage.

Can I guess the voltage across the inductor as I have guessed current at t= 0?

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  • \$\begingroup\$ Current is through, voltage is across. \$\endgroup\$
    – JRE
    Jul 19, 2022 at 13:22

3 Answers 3

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Hint

The steady-state current through the inductor before the switch opens is easy to calculate but, as it happens, is of no consequence because, when the switch opens, that current finds a nice route through the capacitor and, at t=0+, the rate of change of inductor current is basically zero so, what does the inductor formula tell you is the correct answer?


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  • \$\begingroup\$ When the switch is opened why the rate of change of current across inductor will be zero? And if it is so then obviously voltage across inductor will be zero aswell at t+. \$\endgroup\$
    – Alex
    Jul 19, 2022 at 13:43
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    \$\begingroup\$ @Alex at the instant when the switch opens, the capacitor resists any change in its terminal voltage and assists the inductor to maintain its current at a steady value. Of course that current is going to fall immediately after t=0+ but that isn't the question. \$\endgroup\$
    – Andy aka
    Jul 19, 2022 at 14:05
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Intuitively, an inductor resists sudden changes in current by delivering sudden (and for an ideal inductor, arbitrarily large) changes in voltage. A capacitor is the complement: it resists sudden changes in voltage by delivering sudden (and possibly arbitrarily large) changes in current.

So when the switch opens, the capacitor will hold the voltage on the right end of the inductor constant. The inductor will hold its current constant. Because the inductor holds its current constant, the current through the resistor on the left will remain constant, and its voltage will remain constant.

Consequently, the inductor voltage will -- in that instant -- remain constant. Note that the capacitor current will instantaneously jump, taking up all the current that had been flowing in the \$1\Omega\$ resistor.

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Find the steady-state conditions with the switch closed and open, then plug them into the differential equation for the switch-open condition. Namely at \$t=0\$ and \$t \rightarrow \infty\$.

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