Question

An inductance of 1 henry and a resistance of 1 ohm are connected in series with a constant EMF of E volts. If the current is initially zero, and is equal to 10 A after 5 seconds, find E.

Answer 1

$L\frac{dI}{dt}+RI=E$

$1\frac{dI}{dt}+1I=E$

$\frac{dI}{dt}+I=E$

$\frac{dI}{dt}=E-I$

$\frac{dI}{E-I}=dt$

integrate both sides;

$\int \frac{dI}{E-I}=\int dt$

$-log(E-I)|_0^i=t|_0^t$

at t=5, I=10A

Replace the above conditions in the integral;

$-log(E-10)=5$

$log\frac{E}{10}=-5$

$\frac{E}{10}=e^{-5}$

$E=e^{-5}*10$

$E=0.0674V$