Question

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.

Answer 1

Answer:

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))

Answer 2

The mesh law and the solution of the equation for RL circuits allows to find the result for how the current changes in an RL circuit is:

• $i =i_o \ ( 1 - e^{-\frac{R}{L} \ t} )$i

An RL circuit is an electro circuit made up of a power source, a resistor, and an inductor.

If in an electrical circuit the entire circuit runs through the sum of all the voltages must be zero. They indicate the expression for the voltages in the circuit is

           $E - V_R - V_L = 0$  

The voltage across a resistor.

           $V_R$ = i R

The voltage across an inductor.

          $V_L = L \ \frac{di}{dt}$  

Let's substitute.

            $E- i R - L \ \frac{di}{dt}=0$  

Let's solve this expression.  Let's start by separating the variables.

             $\frac{di}{E-iR} = \frac{dt}{L}$

             

Let's  change the variable.

           x = E - iR

           dx = - R di

we substitute

          $- \frac{1}{R} \ \frac{dx}{x} =\frac{dt}{L}$

Let's perform the integral and evaluate from t = 0 to t = t.

         $- \frac{1}{R } \int\limits \, \frac{dx}{x} = \frac{1}{L} \int\limits^t_0 \, dt$  

        $ln X + C = - \frac{R}{L} \ t$

         

we eliminate the natural logarithm.

          $x = e^{- \frac{R}{L} t - C } \\A = e^{-C} \\x = A \ e^{- \frac{R}{L} t }$

       

         E - i R =A $e^{- \frac{R}{L} t$  

To evaluate the constant we use that when the circuit t = 0 is closed the current is zero

           

        E = A

         

We substitute.

         E - iR = E $e^{- \frac{R}{L} t}$  

We clear the expression.

       iR = E - E $E^{- \frac{R}{L} t}$

   

From ohm's law.

        V = iR  

        i = $\frac{V}{R}$  

Let's substitute.

        i = i₀ ( $1 - E^{- \frac{R}{L} \ t} )$  

In conclusion using the mesh law and the solution of the equation we can find the result for how the current changes in an RL circuit is:

• $i = i_o \ ( 1 - e^{- \frac{R}{L} \ y} )$

 

Learn more about RL circuits here:  brainly.com/question/17050299