Now that we have a feel for the state of the circuit in its steady state, let us

ID: 1495735 • Letter: N

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Now that we have a feel for the state of the circuit in its steady state, let us obtain the expression for the current in the circuit as a function of time. Note that we can use the loop rule (going around counterclockwise): To understand the behavior of an inductor in a series R-L circuit. In a circuit containing only resistors, the basic (though not necessarily explicit) assumption is that the current reaches its steady-state value instantly. This is not the Mote as well that v_r = iR and V_i = L. Using these equations, we can get, after some arranging of the variables and making the substitution Integrating both sides of this equation yields Express your answer in terms of e, R, and L. You may or may not need all these variables Use the notation cxp(x) for c^x.

Explanation / Answer

x = xo * exp(-Rt/L)

where, x = E/R - i(t)

At t = 0 , i(0) = io = 0

=> xo = E/R

=> E/R - i(t) = E/R * exp(-Rt/L)

=> i(t) = E/R - E/R * exp(-Rt/L)

=> i(t) = E/R(1 - exp(-Rt/L))

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Now that we have a feel for the state of the circuit in its steady state, let us

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