EDIT:
@egreg 's method works somewhat. As the other equations in the longer code were too long, the whole second column were flushed to the right by these. As I shortened the equations down, the equal signs came closer and closer to the left.
What do I want?:
I am trying to make multiple alignments in my equations using the align enviroment.
Deeper explanation:
I've searched in the forum and tried to use one method of specifying the second column by using double "&" (&&). Also I've tried just using a single "&" (&) a second time within the same equation. But in both of attempts my second alignment makes all the righthand text to be flushed all the way to the right.
This problem occurs when its inbetween some other equations. But when I try it with the double "&" (&&) in code alone, it works pretty good except that it still will be pretty far to the right.
Why I dont want to split the align enviroment into several enviroments:
I want all the equations before and after to also be aligned with these equations in the first column.
Commands I use in the code:
\DeclareMathOperator{\laplace}{\mathcal{L}} %Laplace symbol
\newcommand\ddfrac[2]{\frac{\displaystyle #1}{\displaystyle #2}} % Makes some quotient more visible
Using extra spacing to clearify where I try to make my second alignment.
Here is the code alone, attempt with a double "&" (&&) to specify second column:
\begin{align}
u_{in}(t) &= u_{ut} + u(t)\\
%
\nonumber
i(t) &= i_L(t) + i_R(t)
%
\intertext{Laplace-transformering av ovanstående ekvationer ger:}
%
U_{ut}(s) &= \laplace\left\{ u_{ut}(t) \right\}(s) &&= \frac{1}{Cs}I(s) \label{eq:u_ut}\\
%
U(s) &= \laplace\left\{ u(t) \right\}(s) = L[sI_L(s)-i(0)] &&= sLI_L(s) \label{eq:u}\\
%
U(s) &= \laplace\left\{ \widetilde{u}(t) \right\}(s) &&= RI_R(s) \label{eq:omhs}\\
%
U_{in}(s) &= \laplace\left\{ u_{in}(t) \right\}(s) &&= U_{ut}(s) + U(s) \label{eq:u_in}\\
%
I(s) &= \laplace\left\{ i(t) \right\}(s) &&= I_L(s) + I_R(s) \label{eq:i}\\
%
\intertext{Överföringsfunktionen blir då:}
%
H(s) &= \frac{U_{ut}(s)}{U_{in}(s)}
\end{align}
This gives partly this result:
Here is the full align enviroment, pretty much code but that is why I inserted the example above.
\begin{align}
\nonumber
u_{ut}(t) &= \frac{1}{C}\int_0^ti(t)dt\\
%
\nonumber
u(t) &= L\frac{di_L(t)}{dt}\\
%
\intertext{Ohm's lag:}
%
\nonumber
u(t) &= \widetilde{u}(t) = i_R(t)R\\
%
\nonumber
\intertext{Kirchoff's lagar:}
%
\nonumber
u_{in}(t) &= u_{ut} + u(t)\\
%
\nonumber
i(t) &= i_L(t) + i_R(t)
%
\intertext{Laplace-transformering av ovanstående ekvationer ger:}
%
U_{ut}(s) &= \laplace\left\{ u_{ut}(t) \right\}(s) &&= \frac{1}{Cs}I(s) \label{eq:u_ut}\\
%
U(s) &= \laplace\left\{ u(t) \right\}(s) = L[sI_L(s)-i(0)] &&= sLI_L(s) \label{eq:u}\\
%
U(s) &= \laplace\left\{ \widetilde{u}(t) \right\}(s) &&= RI_R(s) \label{eq:omhs}\\
%
U_{in}(s) &= \laplace\left\{ u_{in}(t) \right\}(s) &&= U_{ut}(s) + U(s) \label{eq:u_in}\\
%
I(s) &= \laplace\left\{ i(t) \right\}(s) &&= I_L(s) + I_R(s) \label{eq:i}\\
%
\intertext{Överföringsfunktionen blir då:}
%
\nonumber
H(s) &= \frac{U_{ut}(s)}{U_{in}(s)} =
%
\intertext{Insättning av ekvationerna \ref{eq:u_ut} \& \ref{eq:u} ger:}
%
&=\frac{I(s)}{sC\left(\frac{1}{sC}I(s)+U(s)\right)} = \ddfrac{I(s)}{I(s)+sCU(s)} \label{eq:h1}\\
%
\intertext{Insättning av ekvation \ref{eq:i} i ekvation \ref{eq:omhs} ger:}
%
\nonumber
U(s) &= RI_R(s) = R(I(s)-I_L(s)) =
%
\intertext{Insättning av ekvation \ref{eq:u} ger:}
%
\nonumber
&= R\left(I(s)-\frac{U(s)}{sL}\right)\\
%
\intertext{Alltså:}
%
\nonumber
U(s) &= R\left(I(s)-\frac{U(s)}{sL}\right)
%
\intertext{Bryt ut U(s):}
%
U(s) &= \ddfrac{R}{\left(1+\frac{R}{sL}\right)}I(s) = \frac{sRL}{R+sL}I(s) \label{eq:u2}
%
\intertext{Insättning av ekvation \ref{eq:u2} i ekvation \ref{eq:h1} ger:}
%
\nonumber
H(s) &= \ddfrac{I(s)}{I(s)+\frac{s^2RLC}{R+sL}I(s)} = \ddfrac{1}{1+\frac{s^2RLC}{R+sL}} = \ddfrac{R+sL}{R+sL+s^2RLC}\\
%
\intertext{Alltså:}
%
H(s) &= \ddfrac{R+sL}{R+sL+s^2RLC} \label{eq:h_laplace}
\end{align}
Which gives me this (the interesting part):
