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I'm not quite sure what the question means. I need to explain what would happen if I removed the \$1.0 \mu F\$ blocking capacitor from this circuit. Howver, I'm not sure what the "DC bias voltage of the emitter" is. Is it simply the voltage as measured from base to emitter? This is for a 2N3904 transistor and an emitter-follower setup.

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They probably mean the voltage on the emitter relative to ground. It should not matter though.

This is a homework questions so I won't give you the exact answer, but think about what happens after a long time has passed to the current through a capacitor and how that would affect the voltage.

If Trevor/KeleMP are correct that they mean to replace it by a short (we can't see the actual wording in the question), the analysis becomes straightforward and you can discover why you would want a blocking capacitor in there. On looking up the actual wording, I think they are correct:

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    \$\begingroup\$ +1 It also depends on your definition of remove.... Can mean short it out or remove it entirely. \$\endgroup\$ – Trevor_G Sep 22 '17 at 19:18
  • \$\begingroup\$ Ok, let's see if I can reason it out. Please help as much as you can. A capacitor stores charge in some time, then it releases it. These time intervals are usually relatively short. However, one thing I don't get is that the capacitor is located "after" the emitter, so how would the existance of the capacitor affect the emitter voltage at all? \$\endgroup\$ – whatwhatwhat Sep 22 '17 at 19:50
  • \$\begingroup\$ You want to determine the current through the capacitor (or the impedance of a capacitor). One formula involves time and one involves frequency. Time is very long for DC and frequency approaches zero. \$\endgroup\$ – Spehro Pefhany Sep 22 '17 at 19:55
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    \$\begingroup\$ I think in removing it the idea is that it would be replaced with a shorting link. The two resistors and their end points would now share your emitter current in interesting ways. \$\endgroup\$ – KalleMP Sep 22 '17 at 21:23
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    \$\begingroup\$ They literally mean to physically remove the capacitor, yes. \$\endgroup\$ – whatwhatwhat Sep 22 '17 at 21:38

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