Transformers and DC voltage Ok, if I have a transformer with 100 coils of the primary and 200 coils on the secondary and I connect a 9 volt battery to the primary, how many volts should I expect from the secondary coil? I know the answer is 0V DC but I need to know why that is the answer.
 A: 
I need to know why that is the answer.

(Jump to the summary if you want to skip the details).
As you may know, a transformer is essentially two inductors that share most if not all of the magnetic flux linkage.
Recall the formula for the voltage across an isolated inductor
$$v_L(t) = L \frac{di_L}{dt}$$
Now consider two inductors, $L_1$ and $L_2$ with perfect magnetic coupling.  The voltage across each is given by:
$$v_1(t) = L_1 \frac{di_1}{dt} + M \frac{di_2}{dt}$$
$$v_2(t) = M \frac{di_1}{dt} + L_2 \frac{di_2}{dt}$$
where
$$M = \sqrt{L_1 L_2}$$
This configuration is essentially a (nearly ideal) transformer.  Let $L_1$ by the primary and $L_2$ be the secondary.
Assume the primary has a constant voltage $V_1$ applied at $t=0$ and the secondary is open (which means that the secondary current is zero).
According to the above, the current in the primary changes at the rate of $\frac{V_1}{L_1}$ amps per second thus the voltage across the secondary is
$$v_2 =  M \frac{di_1}{dt} = \sqrt{L_1 L_2}\frac{V_1}{L_1} = V_1 \sqrt{\frac{L_2}{L_1}}$$
The voltage across the secondary is non-zero and constant!  At least according to this model it is.
What's missing in this analysis?  We haven't taken into account the non-zero resistance of the windings.
Initially, the analysis above holds but, as the primary current increases, the resistance of the primary winding becomes a significant factor.  Instead of a steadily changing current, the primary current is of the form
$$i_1(t) = \frac{V_1}{R_1}\left(1 - e^{-\frac{tR_1}{L_1}} \right)\cdot u(t)$$
where $R_1$ is the resistance of the primary winding and $u(t)$ is the unit step function.
When the voltage is connected to the primary, the primary current initially changes at the maximum rate but rather quickly approaches a constant value.  But the secondary voltage is proportional to the how quickly the primary current changes.
Thus, the voltage across the secondary is
$$v_2(t) = V_1\sqrt{\frac{L_2}{L_1}}e^{-\frac{tR_1}{L_1}}\cdot u(t)$$
In other words, at the instant the voltage is connected to the primary, the secondary voltage jumps to its peak value and decays exponentially to almost zero by time $t = 5\frac{L_1}{R_1}$ seconds.
To summarize, the reason that the secondary voltage is not constant and decays quickly to zero is that the primary current rapidly approaches a constant determined by the applied voltage and the primary winding resistance.
A: The transformer operation is based on Faraday's Law
$$
\nabla\times\mathbf{E} = -\dfrac{\partial\mathbf{B}}{\partial t}
$$
This law relates the generated electric field (resulting in electromotive force in a circuit) with the variation of the magnetic flux density.
Also indicates that a changing electric field generates a density also varying magnetic flux.
Therefore, if a battery to the primary of a transformer is connected, not a variable flux density occurs, and therefore no energy transfer to the secondary winding. This concludes that no voltage is observed in the secondary.
To be more exact, in moments of connection and disconnection of the battery, you can register a voltage in the secondary, but this is only temporary.
A: I'm going to write a really simple answer...
A short summary of what goes on in a step-up transformer:


*

*In an AC source, both the magnitude and direction of current changes

*As the primary windings of the transformer is connected to an AC source, there will be a change in magnetic flux (remember: the magnetic field strength around the coil is dependant on the current though the coil- change in current causes change in magnetic field strength and thus a change in magnetic flux)

*Thus there is a varying magnetic flux in the iron core which in turn causes a varying magnetic flux in the secondary coil

*According to Farady's law, as there is a change in magnetic flux through the secondary winding, there will be an emf induced. (You may think of the coil "cutting" magnetic field lines if that makes it easier to understand)


Now if you think about, its really because of the AC source in the first place that caused the varying magnetic flux which eventually caused the emf to be induced in the secondary coil.
Look at this diagram: (focus on the red graphs)

The first graph shows a DC source. As you can see, the current is unchanged. As magnetic flux density and thus magnetic flux of the coil is dependent on current, no change in current means no varying magnetic field. No change in magnetic flux through the secondary coil means no emf is induced.
In the second graph, a changing current causes a change in magnetic flux and the process continues as stated above.
(I took the image from here)
A: In order for an emf to be induced in the secondary coil, the flux through it must be changing; therefore, the current in the primary coil must also be changing. If a constant voltage is supplied to the primary coil, no emf would be induced in the secondary, and therefore, the secondary voltage would be zero.
