# Imaginary voltage in simple RC circuit

As a homework assignment, we have to find $U_a(\omega)$, that is the voltage that drops over the right resistor in relation to the frequency $\omega$ of the input AC voltage $U_e$.

For the two extreme cases, $\omega \to 0$ and $\omega \to \infty$ I expect $U_a = 0$ since in the $\omega \to 0$ case, the capacitors will block the whole current. In the $\omega \to \infty$ case, the left capacitor will short the whole thing so that the right resistor does not get any current through it.

To get the voltage $U_a(\omega)$ I tried these steps:

• Calculate the total impedance of the circuit $Z$: $$Z(\omega) = R + \left( \frac 1{Z_C(\omega)} + \frac 1{Z_C(\omega) + R} \right)^{-1}$$

The impedance of the capacitor with capacitance $C$ is: $$Z_C(\omega) = \frac{1}{i \omega C}$$

• Calculate the total current: $$I(\omega) = \frac{U_e(\omega)}{Z(\omega)}$$

• Use the current divider rule to get the current that flows through the right branch (with capacitor and resistor). The current should be just this: $$I_r(\omega) = I(\omega) \frac{Z_C(\omega)}{2 Z_C(\omega) + R}$$

• Multiply that current with $R$ and get the result: $$U_a(\omega) = I_r(\omega) R$$

When I plot real and imaginary parts of $U_a$ against $\omega$, I get the following: (The top line is the real part, the bottom line the imaginary part.)

When I plot the absolute value of $U_a$, I get the following plot:

Which one is the current that is really measured? The real part of the absolute value?

I have to find out the value of $\omega$ for which the output voltage is the highest. I tried $\frac{\mathrm d U_a(\omega)}{\mathrm d \omega} = 0$, but I only get $\omega = \pm \frac{i}{RC}$ as solutions, which do not really make sense to me. How can I find the actual maximum?

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Don't forget the context you're working in. You're solving for the phasor voltage across the resistor. When you measure the actual time domain voltage with, say, an oscilloscope, you'll see a sinusoid with an amplitude and a phase (referenced to the source $U_e$).

The magnitude of $U_a$ is the amplitude you will measure. The angle (phase) of $U_a$ is the phase you will measure.

UPDATE: In response to the last question: First, we limit $\omega$ to being a real number. Now, if you work out $\frac{U_a}{U_e}$ by hand (something I highly recommend if you haven't), you should get (assuming I've not made an error):

$\frac{U_a}{U_e} = \dfrac{j \omega RC}{1 - (\omega RC)^2 + j3 \omega RC}$

Now, you can "see" where this is maximum without calculation. As $\omega$ increases from zero, the real part of the denominator is decreasing and becomes zero when $\omega = \frac{1}{RC}$. From that point on, the magnitude of the denominator increases faster than the magnitude of the numerator.

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Okay, so the $|U_a|$ is the actual measured voltage then? – Martin Ueding Jul 1 '12 at 12:53
Could you maybe say something to the question in the last paragraph that I added? – Martin Ueding Jul 1 '12 at 13:01
If you measure the voltage with an AC voltmeter, you'll measure the rms value of the sinusoid. Assuming your $U_e$ is an rms phasor, then yes, $|U_a|$ is the actual measured AC voltage – Alfred Centauri Jul 1 '12 at 13:02
The solution is $\omega = \frac{1}{RC}$. To see this, set $\frac{d|U_a(\omega)|}{d \omega}=0$ – Alfred Centauri Jul 1 '12 at 13:13
When I do that in Mathematica (Solve[D[Abs[u2a[omega]], omega] == 0, omega]), I still the two complex (conjugated) solutions. Am I doing something wrong or is there a trick to it? – Martin Ueding Jul 1 '12 at 13:19