0
\$\begingroup\$

I was reading the Wikipedia page about the applications of capacitor, when I saw an example there for a simple oscillator circuit :

Yet I noticed the capacitor seems to be connected to the (-) terminal from both sides.

How can this circuit function when the capacitor is connected in such a way ? am I missing something here?

\$\endgroup\$
  • 2
    \$\begingroup\$ The capacitor is not connected directly to ground/- on either end - there are other components between the capacitor terminals and ground. \$\endgroup\$ – Peter Bennett Jun 13 '17 at 23:02
  • \$\begingroup\$ @PeterBennett Yes there are, i did not suggest otherwise, since there isn't a clear path that i could notice, but then again, what am i missing ? \$\endgroup\$ – soundslikefiziks Jun 13 '17 at 23:15
  • 1
    \$\begingroup\$ Neither of the two sides of the capacitor are at ground potential (-) \$\endgroup\$ – DerStrom8 Jun 14 '17 at 0:38
  • 1
    \$\begingroup\$ electronics.stackexchange.com/questions/261288/… \$\endgroup\$ – G36 Jun 14 '17 at 3:25
2
\$\begingroup\$

The capacitor is not connected directly to ground. The lamp on one end and the resistor on the other end keep the capacitor above ground potential.

The capacitor is providing feedback from the output to the input. This causes the circuit to oscillate.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.