0
\$\begingroup\$

Here's a simple RC circuit from an AC analysis book:

schematic

simulate this circuit – Schematic created using CircuitLab

I'm trying to determine the phase shift produced by C1 relative to that of V1 given only frequency and impedance. In the book it says that the answer is -25 degrees. Does the negative sign mean that the current lags the source voltage, contrary to the fact that capacitances cause a leading phase shift?

EDIT: Or, does the negative sign signify that it's referring to the actual phase shift minus 90 degrees? Using the tangent relationship between the capacitive reactance and resistance, I keep getting 65 degrees for an answer.

\$\endgroup\$
  • \$\begingroup\$ If I read you correctly, the question is, "What does \$V_C\$ look like relative to \$V_1\$?" If so, then you are asking to compute a voltage divider that looks like \$V_C=V_1\cdot\frac{47\:\textrm{nF}}{47\:\textrm{nF}+1700\: \Omega}\$. The answer to that is \$V_1\$ times \$0.821346633 - 0.383061798\cdot j\$ or \$0.906281763\: \angle -25.003528^\circ\$. \$\endgroup\$ – jonk May 27 '17 at 4:27
  • \$\begingroup\$ Right. Does it mean that the capacitor voltage is lagging behind the source voltage? \$\endgroup\$ – DorkOrc May 27 '17 at 4:33
  • \$\begingroup\$ Yes. The peak voltage measured from \$V_C\$ to ground, relative to \$V_1\$ to ground, will be about \$25^\circ\$ after \$V_1\$'s prior peak. This means about \$\frac{1}{929}\cdot\frac{25}{360}\approx 75\:\mu\textrm{s}\$ later. \$\endgroup\$ – jonk May 27 '17 at 4:42
  • \$\begingroup\$ I see now. I thought the voltage across the cap would behave exactly like in a resistor, where current and voltage are in-phase. Am I right if I say the resistor in the circuit is what introduces the phase shift between the source and the capacitor voltages? \$\endgroup\$ – DorkOrc May 27 '17 at 5:03
  • \$\begingroup\$ If there were no resistor, there'd be no phase shift and the voltage on the cap would track \$V_1\$. This is because a voltage source can supply any amount of needed charge, instantly. But with a resistor in between, the capacitor can only get the current limited by the resistor, so the voltage on the cap is slow to catch up. When \$V_1\$ reaches its peak, the capacitor is still charging upwards and hasn't yet reached \$V_1\$'s voltage. Then, \$V_1\$ falls back downward and at some point there is no voltage across the resistor and the cap voltage stops changing. Then \$V_1\$ drops still more. \$\endgroup\$ – jonk May 27 '17 at 5:08
1
\$\begingroup\$

In the book it says that the answer is -25 degrees

It means that the output sinewave voltage lags the input sinewave voltage by 25 degrees.

Phase lag = \$Tan^{-1}(\dfrac{R}{X_C})\$

Using the tangent relationship between the capacitive reactance and resistance, I keep getting 65 degrees for an answer.

It sounds like you might be taking the arctan of Xc/R (67.08 degrees).

\$\endgroup\$
1
\$\begingroup\$

One way to interpret is the one which you have stated that the waveform is leading. The other way is the one which your calculator is not allowing you to understand, usually in some calculators this issue occurs that if the phase angle goes more than 180 degrees, they just convert it into a negative number. If you add 180+25 degrees to the current phase angle you will see what the actual phase angle is which would be 205 degrees and then you will get what you expect, that is a lagging voltage across the capacitor.

Now... why add 180+(magnitude of the number)? In my calculus class when complex numbers were being taught we used to draw the real and imaginary axis, taking the imaginary part to be negative and real part to be positive that is the vector (in the circuits case which is a phasor) lies in the third quadrant, now if something lies in the third quadrant you can be sure that the angle will be greater than 180 degrees, what the calculator does is it takes the angle from the negative x-axis rather than the positive one and since it does that and is still going counter clockwise, we get the negative sign, so we add 180 to get the angle from the positive x or real axis, which is usually from where it is measured .

Speaking in terms of circuits what happens is, getting -25 degrees does not mean that the voltage has started to lead the entire input, it means that it lags it by 205 degrees, but 205 degrees is such a big lagging (usually anything above 180 degrees is) that we just say that it sort of looks like it is leading, which is actually what is done when we see something having a phase shift of greater than 180 degrees because the waveforms just give a better feel the other way around ie leading instead of lagging because the waveform of the output looks to be coming before the input one. .

Real sorry for not being able to give you the graphical interpretations at the moment which would have given a clearer picture of what is above but if you plot the waveform of the circuit, you will see that the phase difference between the output and input, that is output_angle - input_angle
will be positive if you start from the angle of the input waveform , and will be negative if you start from the angle of the output waveform of the angle

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.