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A V = 82-V source is connected in series with an R = 1.9-kΩ resistor and an L = 34-H inductor and the current is allowed to reach maximum. At time t = 0 a switch is thrown that disconnects the voltage source, but leaves the resistor and the inductor connected in their own circuit. After the current decreases to 51 % of its maximum value, the battery is reconnected into the circuit by reversing the switch.

At what value of the time t, in milliseconds, does the current reach 78 % of its maximum?

1 Answer

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The RL circuit's current decay is exponential, but without precise timing of the battery reconnection or more details, we cannot provide an exact time in milliseconds for the current to reach 78% of the maximum value.

The student's question is related to the behavior of an RL circuit, which consists of a resistor (R) and an inductor (L) in series, when it is being discharged. According to the given information, when the battery is disconnected, the current decreases exponentially. The formula that describes the behavior of the current at any time after the battery is disconnected is given by I = I_0e^{-t/(L/R)}, where I_0 is the initial current, t is time, L is the inductance and R is the resistance.

From the information provided, we deduce that the current decays to 51% of its maximum at an unknown time after the battery is disconnected. When the battery is reconnected at this reduced current level, we are looking for the time it takes for the current to reach 78% of its maximum. However, to precisely calculate this, we need the time at which the battery is reconnected or alternatively, a more detailed explanation of the switching dynamics and initial conditions. Without this information, we cannot provide an exact time in milliseconds for the current to reach 78% of its maximum.

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