I'm very confused about capacitor & resistor in AC current circuits, I followed the equations but still confused, my question is will the voltage & current be decreased ?

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If we applied an AC current 220v/60Hz on point A let's assume R = 1K C = 100nF, What will voltage and current be on point B ?

Please if you don't mind to let me understand the result by running equations

  • 1
    \$\begingroup\$ Please remember that voltage is a differential signal, in other words is a signal that between two points. So if you apply 220V/60Hz at point A are you implying there is some other point C which is the reference point for the 220V/60Hz voltage of the voltage is applied across point A or B \$\endgroup\$
    – Kvegaoro
    Mar 7, 2017 at 18:32
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    \$\begingroup\$ Current is measured in Amps(A), Voltage is measured in Volts(V), so "220v/60Hz" is a voltage, not a current. Voltage is not applied to a single point (which is why its also known as "potential difference"), so your 220V/60Hz must be applied between point A and some other point. \$\endgroup\$
    – brhans
    Mar 7, 2017 at 18:33
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    \$\begingroup\$ "be decreased" relative to what? \$\endgroup\$
    – Andy aka
    Mar 7, 2017 at 18:38
  • \$\begingroup\$ Take a class like AC and DC Fundamentals. \$\endgroup\$ Mar 7, 2017 at 18:42
  • \$\begingroup\$ an RC impedance calculator keisan.casio.com/exec/system/1258032632 \$\endgroup\$ Mar 7, 2017 at 19:13

2 Answers 2



simulate this circuit – Schematic created using CircuitLab

There is a lot to clarify here:

  • This is a series circuit so the current will be the same everywhere
  • You can't have a current at a point, current flows through something. Think of water in a pipe current is the rate the water moves through it. This is not a perfect analogy but should give you the idea
  • Similarly you can't have voltage at a point you have a voltage or potential difference between two points. I know you will be able to point to text books that say the voltage at point 'C' is but that's only a convenience when the reference point can be assumed. Most people here would assume the reference was point 'B' in this circuit as I have marked it with a 'GND' symbol but the correct thing to say would be "the voltage at point 'C' with respect to point 'B' is".

Now if we want to know the current in this circuit we calculate

$$I = \frac{V}{Z}$$

When calculating \$ Z \$ we must take into account that current in a capacitor lags the voltage by \$90^o\$ so it is not just a case of adding them up.

\$ R = 1k \$, \$ X_c = \frac{-j}{2 \cdot \pi \cdot 60 \cdot 100n} \approx -j \cdot 26.525k \$

We can use Pythagoras to find \$ Z = \sqrt{R^2 + X_c^2} \approx 26.54k\$

So the impedance is dominated by the capacitor and the current will be lagging the voltage by almost 90 degrees.

I'll leave the rest as an exercise.


You have no reference points other than A and B, is this where you are applying your AC voltage? If so it will be at source voltage. This is not a purely resistive circuit so basic formulas used with DC will not apply. About the only thing I can tell you with the information given is that voltage will lag current in this type of circuit.


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