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): E−vR−vL=0. Note as well that vR=iR and vL=Ldidt. Using these equations, we can get, after some rearranging of the variables and making the subsitution x=ER−i, dxx=−RLdt. Integrating both sides of this equation yields x=x0e−Rt/L. Use this last expression to obtain an expression for i(t). Remember that x=ER−i and that i0=i(0)=0. Express your answer in terms of E, R, and L. You may or may not need all these variables. Use the notation exp(x) for ex.

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

Explanation:

E−vR−vL=0

E− iR− Ldi/dt = 0

E− iR = Ldi/dt

Separating te variables,

dt/L = di/(E - iR)

Let x = E - iR, so dx = -Rdi and di = -dx/R substituting for x and di we have

dt/L = -dx/Rx

-Rdt/L = dx/x

interating both sides, we have

∫-Rdt/L = ∫dx/x

-Rt/L + C = ㏑x

x = exp(-Rt/L + C)

x = exp(-Rt/L)exp(C)     A = exp(C) we have

x = Aexp(-Rt/L) Substituting x = E - iR we have

E - iR = Aexp(-Rt/L) when t = 0, i(0) = 0. So

E - i(0)R = Aexp(-R×0/L)

E - 0 = Aexp(0) = A × 1

E = A

So,

E - i(t)R = Eexp(-Rt/L)

i(t)R = E - Eexp(-Rt/L)

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

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

i dont know but

explanation:

whoever your teacher is, deserves a raise for using pokemon as an example

the answer is ampere

the ampere is used to measure electricity. an ampere is a unit of electric current equal. a unit of measure for the flow of current in a circuit. one ampere is the amount of current flow provided when one volt of electrical pressure is applied against one ohm of resistance.