Problem evaluating the current function over a time interval in circuit with inductance

Question

The given expression describes a piecewise function of voltage u(t) across a circuit element with respect to time and is defined over four intervals:

$$ u(t)= \begin{cases} V_{0} \frac{t}{(2T)} &\text{for \(0 \leq t \leq 2T)\)}\\ V_{0} &\text{for \(2T \leq t \leq 3T)\)}\\ -V_{0} &\text{for \(3T \leq t \leq 4T)\)}\\ \frac{V_{0}}{2}(\frac{t}{T}-6) &\text{for \(4T \leq t \leq 8T)\)} \end{cases} $$

for $$0 < t \leq 2T$$

The expression $$i(t)=i(0)+\frac{1}{L} \int_{0}^{t} V_{0}\frac{t}{2T}dt$$ is a relationship between the current i(t) and the voltage V_0 over time for an inductor with inductance L.

Doing the math (if I am correct) I take:

\begin{align*} \frac{1}{L} &* \int_{0}^{t} V_{0}\frac{t}{2T}dt\\ &=\frac{V_{0}}{L} \int_{0}^{t} \frac{t}{2T}dt\\ &=\frac{V_{0}}{L}\dfrac{t^2}{4T}\\ &=\frac{V_{0}}{L}(\dfrac{t}{2T})^2 \end{align*}

the book that I am reading gives the result $$i(t) =\frac{V_{0}T}{L}(\dfrac{t}{2T})^2 $$.

2 questions :

1) First of all the i(0) = ?

2) where the T in the numerator came from in the resulted expression ?

Answer 1

Well, mathematically speaking let's define the voltage in a possible correct way:

$$\text{V}\left(t\right):=\begin{cases} \displaystyle\text{V}_0\cdot\frac{t}{2\text{T}}&\text{if}\space0\leq t<2\text{T}\\ \\ \displaystyle\text{V}_0&\text{if}\space2\text{T}\leq t<3\text{T}\\ \\ \displaystyle-\text{V}_0&\text{if}\space3\text{T}\leq t<4\text{T}\\ \\ \displaystyle\frac{\text{V}_0}{2}\cdot\left(\frac{t}{\text{T}}-6\right)&\text{if}\space4\text{T}\leq t\leq8\text{T} \end{cases}\tag1$$

Now, the relationship between the voltage across and the current through an inductor is given by:

$$\text{V}_\text{L}\left(t\right)=\text{I}_\text{L}'\left(t\right)\cdot\text{L}\space\Longleftrightarrow\space\text{I}_\text{L}\left(t\right)=\text{C}+\frac{1}{\text{L}}\int\text{V}_\text{L}\left(t\right)\space\text{d}t\tag2$$

So, we get:

\begin{equation} \begin{split} \text{I}_\text{L}\left(t\right)&=\text{C}+\frac{1}{\text{L}}\int\text{V}_\text{L}\left(t\right)\space\text{d}t\\ \\ &=\text{C}+\frac{1}{\text{L}}\int\text{V}\left(t\right)\space\text{d}t\\ \\ &=\text{C}+\frac{1}{\text{L}}\cdot\begin{cases} \displaystyle\int\text{V}_0\cdot\frac{t}{2\text{T}}\space\text{d}t&\text{if}\space0\leq t<2\text{T}\\ \\ \displaystyle\int\text{V}_0\space\text{d}t&\text{if}\space2\text{T}\leq t<3\text{T}\\ \\ \displaystyle\int-\text{V}_0\space\text{d}t&\text{if}\space3\text{T}\leq t<4\text{T}\\ \\ \displaystyle\int\frac{\text{V}_0}{2}\cdot\left(\frac{t}{\text{T}}-6\right)\space\text{d}t&\text{if}\space4\text{T}\leq t\leq8\text{T} \end{cases}\\ \\ &=\text{C}+\frac{1}{\text{L}}\cdot\begin{cases} \displaystyle\frac{\text{V}_0}{2\text{T}}\int t\space\text{d}t&\text{if}\space0\leq t<2\text{T}\\ \\ \displaystyle\text{V}_0\int1\space\text{d}t&\text{if}\space2\text{T}\leq t<3\text{T}\\ \\ \displaystyle-\text{V}_0\int1\space\text{d}t&\text{if}\space3\text{T}\leq t<4\text{T}\\ \\ \displaystyle\frac{\text{V}_0}{2}\cdot\left(\frac{1}{\text{T}}\int t\space\text{d}t-6\int1\space\text{d}t\right)&\text{if}\space4\text{T}\leq t\leq8\text{T} \end{cases}\\ \\ &=\text{C}+\frac{1}{\text{L}}\cdot\begin{cases} \displaystyle\frac{\text{V}_0}{2\text{T}}\cdot\frac{t^2}{2}+\text{n}_1&\text{if}\space0\leq t<2\text{T}\\ \\ \displaystyle\text{V}_0t+\text{n}_2&\text{if}\space2\text{T}\leq t<3\text{T}\\ \\ \displaystyle\text{n}_3-\text{V}_0t&\text{if}\space3\text{T}\leq t<4\text{T}\\ \\ \displaystyle\frac{\text{V}_0}{2}\cdot\left(\frac{1}{\text{T}}\cdot\frac{t^2}{2}-6t\right)+\text{n}_4&\text{if}\space4\text{T}\leq t\leq8\text{T} \end{cases} \end{split}\tag3 \end{equation}