Answer to Question #306156 in Differential Equations for Rex

Question

An inductance of L Henrys and a resistance of 10 ohms are connected in series with e.m.f. of 100 volts. If the current is initially zero, and is equal to 9 amperes after 1 second , find L and find the current after 0.5 second

Answer 1

The instantenous emf equation is given as;

$L\frac{di}{dt}+Ri=V$

$\frac{di}{dt}=\frac{V-Ri}{L}$

$\frac{di}{dt}=\frac{R}{L}(\frac{V}{R}-i)$

$\frac{di}{(\frac{V}{R}-i)}=\frac{R}{L}dt$

Integrating on both sides

$\int_0^i \frac{di}{(\frac{V}{R}-i)}=\int_0^t \frac{R}{L}dt$

$\int_0^i \frac{di}{(\frac{i-V}{R})}=-\frac{R}{L}\int_0^tdt$

$ln\frac{(i-\frac{V}{R})}{-\frac{V}{R}}=-\frac{R}{L}t$

$\frac{(i-\frac{V}{R})}{-\frac{V}{R}}=e^{-\frac{R}{L}t}$

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

When t=1, i(1)=9

$9={\frac{100}{10}}(1-e^{-\frac{10}{L}1})$

$\frac{9}{10}=1-e^{\frac{-10}{L}}$

$e^\frac{-10}{L}=0.1$

$\frac{-10}{L}=ln(0.1)$

ln(0.1)=-2.302

$L=\frac{10}{2.302}$

L=4.443H

Current after 0.5s is;

$i(0.5)={\frac{100}{10}}(1-e^{-\frac{10}{4.344}*0.5})$

=6.837A