Engineering Calculus Problem

Question

If $L\frac{di}{dt}+Ri=E $, where $L, R$ and $E$ are constants, and it is known that $i=0$ at $t=0$, show that: $$ \int_{0}^{t} (Ei-Ri^2)dt = \frac{Li^2}{2} \tag1 $$

I'm very frustrated with this. First i noticed that there is no $t$ term appearing anywhere. Only $i$ is a function of $t$, i.e. $i(t)$. So i found that $i(t)=\frac{E}{R}-\frac{L}{R}\frac{di}{dt}$. I took the LHS of $(1)$ and tried to get to the second with no success.

Then i thought of starting from the RHS and trying to prove that $ \frac{L}{2}\frac{di^2}{dt} = Ei-Ri^2 $.

Everything i tried was to no avail. And many questions have arisen along the way. Can someone help me?

Answer 1

HINT:

$$f(t)\frac{df(t)}{dt}=\frac12 \frac{df^2(t)}{dt}$$

Answer 2

Notice, we have the following differential equation $$L\frac{di}{dt}+Ri=E$$ multiplying by $i$, we get

$$Li\frac{di}{dt}=Ei-Ri^2$$ $$Ei-Ri^2=Li\frac{di}{dt}$$
$$(Ei-Ri^2)dt=Li\ di$$
Integrating the above equation & applying proper limits of $t$ from $0\to t$ & $i$ from $0\to i$, we get

$$\int_{0}^{t}(Ei-Ri^2)\ dt= L \int_{0}^{i}i\ di$$ $$= L \left[\frac{i^2}{2}\right]_{0}^{i}=\frac{1}{2}Li^2$$ Hence, we get $$\color{blue}{\int_{0}^{t}(Ei-Ri^2)\ dt=\frac{Li^2}{2}}$$

Answer 3

$$\frac {d}{dt} \int_0^t (E i(x)-R i(x)^2)dx=E i(t)-R i(t)^2=(E-R i(t)) i(t).$$ $$ \text {And }\frac {d}{dt} \frac {L i(t)^2}{2}=L i(t) i'(t)=(L i'(t)) i(t)=(E-R i(t)) i(t).$$ So the LHS and RHS of your Equation (1) differ by a constant,which is seen to be $0$ by the condition $i(t)=0.$