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I have a circuit with a power source and wave generator connected to capacitor connected to resistor. With an oscilloscope probe attached between the capacitor and resistor, why is it that when applying a DC offset there is no visible offset on the oscilloscope?

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    \$\begingroup\$ Is your oscilloscope in DC or AC coupled mode? \$\endgroup\$ – Anindo Ghosh Mar 22 '13 at 8:00
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    \$\begingroup\$ And if it is in DC mode, how are the generator, capacitor and resistor wired up? There should be an "add circuit" button to let you draw a schematic. \$\endgroup\$ – user_1818839 Mar 22 '13 at 10:53
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You have constructed a simple form of high-pass filter. The impedance of the capacitor rises with decreasing frequency, and at DC, it is infinite. Which makes sense, since a capacitor is essentially two metal sheets separated by an insulator — it clearly isn't going to pass a a DC current.

For AC frequencies, the impedance of the capacitor forms a voltage divider with the resistor. There is a frequency at which they are equal, and this is known as the "cutoff frequency" of the filter. Below this frequency, the output level of the filter drops approximately linearly with frequency. Above this frequency, the output level gradually approaches the input level.

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