Understanding this AC coupled circuit
For DC you only have ordinary voltage divider \$Vo = 10V*\frac{2k}{2k+2k} = 5V\$. And now if we connect our capacitor into the circuit the capacitor will charged to 5V after $$T &gt;5RC = 1\mu F * 1k\Omega = 5ms $$  

Why \$1k\Omega \:?\$  What equivalent resistance is seen at the terminals of the capacitor? 

Interesting things will start to happen if we connect AC signal source. 
But you must remember one thing that the current in the capacitor is proportional the rate of voltage change across it (proportional to how quickly the voltage across the capacitor is changing). The faster the voltage change (frequency of an AC signal is high) the large the current flow through the capacitor. 

So if AC signal frequency is high enough (\$F_{sig} &gt; \frac{1}{2 \pi RC} = 160Hz\$) the capacitor will act like a short in the circuit. The capacitor is sufficiently large value (for the frequency of AC signal) so that it does not have time to charge or discharge. That means that voltage across capacitor will be some average \$ 5V \$. And the situation will look like this if AC signal is a sine wave (Fsing > 160Hz and 1Vpeak).



Now the input AC voltage will change the voltage at the voltage divider output from 4V to 6V in "rhythm" of input AC voltage. Because voltage across capacitor is fixed (5V), because it does not have time to charge or discharge, we can consider that (at least short term) it acts as if it was a battery (whose voltage we fixed (5V) by choosing values of voltage divider).