Question

An electromotive force E(t) = 200, 0 ≤ t ≤ 50 0, t > 50 is applied to an LR-series circuit in which the inductance is 50 henries and the resistance is 5 ohms. Find the current i(t) if i(0) = 0.

Answer 1

The answer is below

Step-by-step explanation:

An LR series circuit has a differential equation in the form of:

`L(di)/(dt)+iR=E(t)\\ \\Given\ that\ 50H,R=5\ ohms,E(t)=200,for\ 0 \leq t \leq\ 50. Hence:\\\\50(di)/(dt)+5i =200\\\\(di)/(dt)+0.1i =4\\\\Solving\ the\ differential\ equation:\\\\The\ integrating\ factor(I)=e^{\int\limits {0.1} \, dt }=e^(0.1t). The\ DE\ becomes:\\\\e^(0.1t)(di)/(dt)+e^(0.1t)(0.1i) =4e^(0.1t)\\\\e^(0.1t)i=\int\limits {4e^(0.1t)} \, dt \\\\e^(0.1t)i=40e^(0.1t)+A\\\\i(t)=40+Ae^(-0.1t)\\\\but\ i(0)=0\\\\0=40+Ae^(-0.1(0))\\\\A=-40\\\\i(t)=40-40e^(-0.1t)\\\\`

At 50 seconds:

`i(50)=40-40e^(-0.1*50)\\\\i(50)=40-40e^(-5)`

`L(di)/(dt)+iR=E(t)\\ \\Given\ that\ 50H,R=5\ ohms,E(t)=0,for\ t> 50. Hence:\\\\50(di)/(dt)+5i =0\\\\(di)/(dt)+0.1i =0\\\\(di)/(dt)=-0.1i\\\\(di)/(i)=-0.1dt\\\\\int\limits {(di)/(i)} =\int\limits {-0.1} \, dt\\ \\ln(i)=-0.1t+A\\\\taking\ exponential:\\\\i=e^(-0.1t+A)\\\\i=e^(-0.1t)e^A\\\\i(t)=Ce^(-0.1t)\\\\i(50)=40-40e^(-5)=Ce^(-5)\\\\C=40(e^5-1)\\\\i(t)=40(e^5-1)e^(-0.1t)\\\\`

`i(t)=\left \{ {{40-40e^(-0.1t)\ \ \ \ 0 \leq t \leq 50 } \atop {40(e^5-1)e^(-0.1t)\ \ \ \ t>50}} \right.`