bhoot said:
So what exactly does it mean that when you bring up voltage with a transformer, you lose current?
A transformer works according to a principle of mutual inductance. A changing current in the primary coil induces an emf in the secondary coil, and vice versa. There is still self inductance present, creating a "back emf" from the changing current in its own coil. If you attach a source of emf in the primary and an active load (resistor) in the secondary, then the equations become:
<br />
\begin{array}{l}<br />
\mathcal{E}(t) - L_1 \, \frac{d I_1}{d t} + M \, \frac{d I_2}{d t} = 0 \\<br />
<br />
-L_2 \, \frac{d I_2}{d t} + M \, \frac{d I_1}{d t} = R \, I_2<br />
\end{array}<br />
Assuming harmonic time dependence, these differential equations become ordinary algebraic equations:
<br />
\begin{array}{l}<br />
-i \, \omega \, L_1 \, I_1 + i \, \omega \, M \, I_2 = \mathcal{E} \\<br />
<br />
-i \, \omega \, M \, I_1 + (i \, \omega \, L_2 - R) \, I_2 = 0<br />
\end{array}<br />
The determinant of this system is:
<br />
\Delta = \left\vert \begin{array}{cc}<br />
-i \, \omega \, L_1 & i \, \omega \, M \\<br />
<br />
-i \, \omega \, M & i \, \omega \, L_2 - R<br />
\end{array} \right\vert = \omega^2 \, (L_1 \, L_2 - M^2) + i \, R \, L_1 \, \omega<br />
and the auxiliary determinants are:
<br />
\Delta_1 = \left\vert \begin{array}{cc}<br />
\mathcal{E} & i \, \omega \, M \\<br />
0 & i \, \omega \, L_2 - R<br />
\end{array} \right\vert = (i \, \omega \, L_2 - R) \, \mathcal{E}<br />
<br />
\Delta_2 = \left\vert \begin{array}{cc}<br />
-i \, \omega \, L_1 & \mathcal{E} \\<br />
<br />
-i \, \omega \, M & 0<br />
\end{array} \right\vert = i \, \omega \, M \, \mathcal{E}<br />
so the currents are:
<br />
\begin{array}{l}<br />
I_1 = \frac{(i \, \omega \, L_2 - R) \, \mathcal{E}}{\omega^2 \, (L_1 \, L_2 - M^2) + i \, R \, L_1 \, \omega} \\<br />
<br />
I_2 = \frac{i \, \omega \, M \, \mathcal{E}}{\omega^2 \, (L_1 \, L_2 - M^2) + i \, R \, L_1 \, \omega}<br />
\end{array}<br />
For ideal transformers:
<br />
L_1 \propto N^2_1, L_2 \propto N^2_2, \, M \propto N_1 \, N_2 \Rightarrow L_1 \, L_2 = M^2<br />
and the above expressions simplify to:
<br />
\begin{array}{l}<br />
I_1 = \frac{\mathcal{E}}{R} \, \left( \frac{N^2_2}{N^2_1} + i \, \frac{R}{\omega \, L_1} \right) \\<br />
<br />
I_2 = \frac{\mathcal{E}}{R} \, \frac{N_2}{N_1}<br />
\end{array}<br />
As you may see, the real parts of these currents are in a ratio:
<br />
\frac{I'_1}{I'_2} = \frac{N_2}{N_1}<br />
The voltage drop, on the other hand, is given by:
<br />
V_1 = L_1 \, \frac{dI_1}{d t} - M \, \frac{d I_2}{d t} \rightarrow -i \, \omega \, (L_1 \, I_1 - M \, I_2)<br />
<br />
V_1 = \mathcal{E}<br />
<br />
V_2 = L_2 \, \frac{d I_2}{d t} - M \, \frac{d I_1}{d t} \rightarrow - i \omega \, (L_2 \, I_2 - M \, I_1)<br />
<br />
V_2 = \mathcal{E} \, \frac{N_2}{N_1}<br />
As you can see, the ratio of the voltage drops is:
<br />
\frac{V_1}{V_2} = \frac{N_1}{N_2}<br />
which is an inverse relation to the one obeyed by the ratio of the active parts of the currents.
The voltage drop on the resistor load is also:
<br />
V_R = R \, I_2 = \mathcal{E} \, \frac{N_2}{N_1}<br />
