# AC/Phasor Circuit Output Voltage Question

So we have a pretty simple circuit, just a 10V voltage source in series with an impedance of 1.8k ohms and a capacitor with a capacitance of 100nF. We're asked to check the voltage across the capacitor at various frequencies.

I was told we use the reactance of the capacitor to find the voltage, which is 1/wC. I do (what i thought was) a simple voltage division across the capacitor, which I thought was just V * Xc/(Xc + R) with Xc being the reactance of the capacitor and R being the impedance of the other element.

The answers I get make sense but the solutions say that I should have used this formula: V * Xc / sqrt(Xc^2 + R^2). the answers I get using either formula are both similar but why would I use that second formula? Where'd they get that from?

For a shared current into a resistor and capacitor you might be tempted to say: -

$$\V_{SUPPLY} = V_R + V_C\$$ (incorrect)

This would not be true because the voltage across a capacitor does not rise and fall sinusoidally as the current rises and falls sinusoidally. For a capacitor the current and voltage looks like this: -

(source: electronics-tutorials.ws)

In other words it is 90 degrees out of phase with voltage. This is because the basic formula for a capacitor is

$$\I = C\dfrac{dV}{dt}\$$

And, if V is a sinewave voltage then I has to be a cosine current.

If instead of real waveforms we drew them as phasors we would represent the voltages and current like so: -

So now if we want to "relate" Vsupply to the individual voltages of the capacitor and resistor we have to add them using pythagorous i.e.

$$\V_{SUPPLY} = \sqrt{V_R^2 + V_C^2}\$$.

It follows from this that impedances also add this way.

Let $Z = \sqrt {R^2 + X_C^2}$. This means Z is the hypotenuse of a right triangle, where $X_C$ is the opposite leg.

So $\frac {X_C} {\sqrt {R^2 + X_C^2}}$ is $\frac {opp} {hyp}$, or the sine of a right triangle. If we multiply it by the hypotenuse of another right triangle $V_S$ (applied source voltage), we will get the opposite component of the second triangle or $V_C$.

Kirchhoff's Voltage Law must be satisfied for any circuit.

In a DC circuit, or purely resistive, capacitive or inductive (ideal) AC circuit, your voltage division approach for series circuits would work because all the voltages are going in the same direction.

KVL: $V_S = V_1 + V_2$

But it does not work when different components are connected in an AC circuit.

KVL still works in any AC circuit, but as vectors.

KVL: $\overrightarrow{V_S} = \overrightarrow {V_R} + \overrightarrow {V_C}$

This gets simplified, because in a series circuit the current is in phase with $V_R$ (horizontal) and current leads $V_C$ by 90° (vertical). This forms two legs of a right triangle called the phasor diagram.

KVL: $\overrightarrow{V_S} = \overrightarrow {V_R} + \overrightarrow {V_C} = \sqrt {V_R^2 + V_C^2}$

Opposition forms another right triangle called the impedance triangle $Z = \sqrt {R^2 + X_C^2}$. Power forms a third right triangle called the power triangle $S = \sqrt {P^2 + Q_C^2}$.

Your solutions may look like voltage division, but it really is just trig.