Answer:
[tex]i(t)=(2/3)(1-e^{-300t})[/tex]
Step-by-step explanation:
Before we even begin it would be very helpful to draw out a simple layout of the circuit. Then we go ahead and apply kirchoffs second law(sum of voltages around a loop must be zero) on the circuit and we obtain the following differential equation,
[tex]-V +Ldi/dt+Ri=0[/tex]
where V is the electromotive force applied to the LR series circuit, Ldi/dt is the voltage drop across the inductor and Ri is the voltage drop across the resistor. we can re write the equation as,
[tex]di/dt+Ri/L=V/L[/tex]
Then we first solve for the homogeneous part given by,
[tex]di/dt+Ri/L=0[/tex]
we obtain,
[tex]i(t)_{h} =I_{max}e^{-Rt/L}[/tex]
This is only the solution to the homogeneous part, The final solution would be given by,
[tex]i(t)=i(t)_{h} +c[/tex]
where c is some constant, we added this because the right side of the primary differential equation has a constant term given by V/R. We put this in the main differential equation and obtain the value of c as c=V/R by comparing the constants on both sides.if we put in our initial condition of i(0)=0, we obtain [tex]I_{max} =V/R[/tex], so the overall equation becomes,
[tex]I(t)=(V/R)(1-e^{-Rt/L})[/tex]
where if we just plug in the values given in the question we obtain the answer given below,
[tex]i(t)=(2/3)(1-e^{-300t})[/tex]
