Question: I'm having a little trouble understanding this circuit and could use some help.

So far I understand that this is an AC coupled circuit, and the capacitor is a blocking capacitor. I also know that this is a high pass filter, but if I was to find the cutoff frequency, what resistance would I use? Would I simply use the 2k lower resistor or combine both resistors for a 4k resistor here?

Also what does this circuit do to an input signal? Is everything offset by +10V here, or does the capacitor block this so any signal from V(t) have an average voltage centered around 0?

[![enter image description here](https://i.stack.imgur.com/1fGIX.png)](https://i.stack.imgur.com/1fGIX.png)

Answer: 
**DC circuit:**

\$V\_{out}\$ is just the output of a voltage divider, so you have 5V out. The \$1 \mu F\$ capacitor behaves as an open circuit, so you can ignore V(t) because it's unconnected.

**AC circuit:**

The 10V DC voltage source is shorted and the 1uF capacitor behaves like a "wire" (short-circuit or low impedance). So you have \$V(t)\$ in parallel with a 1k resistor: \$V\_{out} = V(t)\$.

Thus, when considering DC+AC:

$$
V\_{out} = 5V + V(t)
$$

**EDIT:**

For the expression above to be an approximation good enough, you must check that the frequency of \$V(t)\$ is well above the cut-off frequency of the high pass section formed by the capacitor and the parallel of the 2k resistors, thus:

$$
f \gg \frac {1} {2 \pi \cdot 1000 \cdot 10^{-6}} \approx 159.15 Hz
$$

(thanks go to Vladimir for pointing this to me in the comments)
Yes you combine the resistors but you combine them as parallel components hence, as far as the filter is concerned the value of resistance is 1 kohm. The capacitor blocks the dc voltages from getting onto the input. The average value of Vout is 5 volts.

It may not seem obvious as to why you parallel the resistors but, if you imagine the 10 volt dc source as a short circuit to ac signals it should become clear.