If the Switch Is Thrown Open After the Current in an RL

question 6

Multiple Choice

If the switch is thrown open after the current in an RL circuit has built up to its steady-state value, the decaying current obeys the equation $\mathrm { L } \frac { \mathrm { di } } { \mathrm { dt } } + \mathrm { Ri } = 0$ . How long after the switch is thrown open will it take the current to fall to $40 \%$ of its original value?

• A) \(- 4.60 \mathrm {~L} / \mathrm { R }\) seconds
• B) \(0.92 \mathrm {~L} / \mathrm { R }\) seconds
• C) \(1.12 \mathrm {~L} / \mathrm { R }\) seconds
• D) \(- 4.40 \mathrm {~L} / \mathrm { R }\) seconds

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