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In the following circuit, does V1=V2? It seems to be the case, because there is nothing being driven across the capacitor terminals.

enter image description here

The relevant application is filtering the signal received from a pickup coil, which I guess would be done by a capacitor in parallel.

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    \$\begingroup\$ V1 = V2 until you connect a load to V2. Then there will be a phase shift and some attenuation (a high-pass filter). \$\endgroup\$
    – Transistor
    Jun 13 '17 at 22:03
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Since the voltage source for the circuit is AC, the capacitor will exhibit reactance. This reactance is in series with the source so it can have an effect on the output voltage depending on the load placed at the output.

The reactance of the capacitor is frequency dependent so if the frequency range of the AC signal is large enough the capacitor, along with any series resistance, acts as a filter. That probably will not be the case in your application.

The capacitor also blocks any DC component that might be present with the AC signal. This is a common decoupling technique to yield only the desired AC signal.

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  • \$\begingroup\$ There isn't anything being driven across the capacitor though, so wouldn't the 'ground' terminals of V1 and V2 be the same, and thus no voltage across the capacitor? The capacitor is essentially an appendage with no return path for current to flow. \$\endgroup\$
    – abc
    Jun 14 '17 at 0:42
  • \$\begingroup\$ A ground connection is a relative term. Whether the capacitor is placed in series with the positive side or the negative side makes no difference. Now if you reference the lower V1 connection to ground on the same circuit with which you are trying to measure V2, then it will not work since the cap is then shorted out. But with V1's connections open circuited, the cap will function as described. \$\endgroup\$
    – Glenn W9IQ
    Jun 14 '17 at 1:07

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