Consider an L-R circuit as shown in the figure. The battery provides 12.0 of vol

Question

Consider an L-R circuit as shown in the figure. The battery provides 12.0 of voltage. The inductor has inductance , and the resistor has resistance = 150 . The switch is initially open as shown. At time , the switch is closed. At time after the current flows through the circuit as indicated in the figure.

After the switch is closed, the current in the circuit grows over time approaching a constant value. In general, at time after a voltage source is connected to an L-R circuit, the current in the circuit is given by the expression

I(t) = (V/R)(1-e^(-t/(time constant))

where is the voltage provided by the battery, R is the resistance of the resistor, and time constant is the time constant characteristic of the circuit.

What is the current flowing in the circuit shown in the figure at one time constant after ?

Explanation / Answer

Given that in LR circuit a battery V = 12.0 V Resistance R = 150 To find the current in the circuit after one time constant, I ( t ) = V / R ( 1 - e^ -1 / 1 )            = 12.0 V / 150 ( 1- e^-1 )            = 5.06*10^-2 A

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