Learning Goal: To understand the mathematics of current decay in an L-R circuit

Question

Learning Goal: To understand the mathematics of current decay in an L-R circuit Learning Goal: To understand the mathematics of current decay in an L-R circuit A DC voltage source is connected to a resistor of resistance R and an inductor with inductance L, forming the circuit shown in the figure. For a long time before t=0, the switch has been in the position shown, so that a current I_0 has been built up in the circuit by the voltage source. At t=0 the switch is thrown to remove the voltage source from the circuit. This problem concerns the behavior of the current I(t) through the inductor and the voltage V(t) across the inductor at time t after t=0.

Explanation / Answer

Current decay in an R-L circuit
Now suppose switch S1 in the circuit of figure has been closed for a while and that the current has reached the value Io.Resetting our stopwatch to redefine the initial time,we close switch S2 at time t = 0,bypassing the battery.The current through R and L does not instantaneously go to zero but decays smoothly,as shown in figure.
The Kirchoff's-rule loop equation is obtained from equation
E - iR - L * (di/dt) = 0 ---------(1)
by simply omitting the term.The current i varies with time according to
i = Io * e^-(R/L)t ----------(2)
where Io is the initial current at time t = 0.The time constant,t = (L/R),is the time for current to decrease to 1/e,or about 37%,of its original value.In time 2t it has dropped to 13.5%,in time 5t to 0.67%,and in 10t to 0.0045%.
The energy that is needed to maintain the current during this decay is provided by the energy stored in the magnetic field of the inductor.We can write in the equation form as 0 = i^2R + Li * (di/dt) In this case,Li * (di/dt) is negative.The above equation shows that the energy stored in the inductor decreases at a rate equal to the rate of dissipation of energy i^2R in the resistor.
For a long time before , the switch has been in the position shown, so that a current has been built up in the circuit by the voltage source. At the switch is thrown to remove the voltage source from the circuit. This problem concerns the behavior of the current through the inductor and the voltage across the inductor at time after .

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