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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?

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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: -

http://www.electronics-tutorials.ws/capacitor/cap31.gif?81223b

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: -

enter image description here

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.

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Simple Answer: Right triangles.

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\$.

Long answer:

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.

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