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

Answer:

The answer is below

Explanation:

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

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

At 50 seconds:

$i(50)=40-40e^{-0.1*50}\\\\i(50)=40-40e^{-5}$

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

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