Equal Peak Voltages in a Circuit: Is It Always True?

[Frigus]

In highlighted statement they are saying that vector sum of peak voltages of all components in circuit is equal to peak voltage of battery,I think they are saying this due to the fact that vertical components of both b and c diagrams are equal but what if their horizontal components are not equal?

 

[etotheipi]

Well, they made a pretty bad mistake anyway. They keep saying they "combined" $V_L \in \mathbb{C}$ and $V_C \in \mathbb{C}$ into $V_L - V_C$, but what they are referring to should of course actually be $V_L + V_C$.

 

[Frigus]

Well, they made a pretty bad mistake anyway. They keep saying they "combined" $V_L \in \mathbb{C}$ and $V_C \in \mathbb{C}$ into $V_L - V_C$, but this should of course actually be $V_L + V_C$.

Aren't they right because most of the times voltages across inductor and capacitor have opposite polarity?

 

[hutchphd]

Well, they made a pretty bad mistake anyway. They keep saying they "combined" $V_L \in \mathbb{C}$ and $V_C \in \mathbb{C}$ into $V_L - V_C$, but what they are referring to should of course actually be $V_L + V_C$.

I can't really see all the text but when doing the weird "phasor" stuff it is common to define the voltages to be the length of the phasor and therefore positive. So not an absolute mistake
Please just teach complex numbers (this looks like Sears Zemansky et al)

 

[etotheipi]

I can't really see all the text but when doing the weird "phasor" stuff it is common to define the voltages to be the length of the phasor and therefore positive. So not an absolute mistake
Please just teach complex numbers (this looks like Sears Zemansky et al)

They explicitly defined $V_{C}, V_{L}$ and $V_{R}$ as phasors, and even refer to their vector sum. So I think, this is an absolute mistake!

Anyway, @Hemant, to answer your question... a phasor is just a complex amplitude $V\text{exp}(i \varphi_0)$ multiplied by a complex phase $\text{exp}(i \omega t)$. Typically, you define them so that the real part is the meaningful part physically, so a priori from Kirchoff you know that the horizontal components are contrained to sum to $\mathscr{E}$. Further, the imaginary part of $\mathscr{E}_m$ is by definition the sum of the imaginary parts of $V_L$, $V_C$ and $V_R$, so it holds trivially on those vertical components.

I don't know if that answers your question?

 

[Frigus]

I can't really see all the text

This is Resnick and Halliday.

 

[hutchphd]

@etotheipi is trivially (but fundamentally) incorrect here. The definition of phasors differs from the normal definition of complex amplitudes so Halliday et al is correct.

 

[etotheipi]

@etotheipi is trivially (but fundamentally) incorrect here. The definition of phasors differs from the normal definition of complex amplitudes so Halliday et al is correct.

No, they are incorrect on the $V_L - V_C$ business. I don't understand why you choose to pick a fight with me over this? Because $\mathbb{C} \cong \mathbb{R}^2$ this really is just vector addition, and they got it wrong.

 

[Frigus]

Please use less mathematics,
I am not studying maths and I now can't afford to study it as I have to study other things too.

 

[hutchphd]

They explicitly defined VC,VL and VR as phasors, and even refer to their vector sum. So I think, this is an absolute mistake!

I am not questioning the physics. The point is that phasors use a stupid convention for the signs and they (Halliday) use it correctly
Do you think that this 50 year old text would still have such an egregious mistake?
The book is correct within their definitions. I don't know why they don't actually just teach Complex Impedance but they don't.

 

[vanhees71]

I don't know, what they mean by phasors. The usual way is to describe a circuit in the stationary state by complex currents and voltages (understanding that the physical signal is its real part).

Now take as an example the series circuit with a resistor $R$ and (ideal) inductor $L$, and a capacitor $C$. Now you take the external source as $U(t)=U_0 \exp(\mathrm{i} \omega t)$ and consider the quasistationary state, where also the current through the resistor, is of the form $i(t)=U_0 \exp(\mathrm{i} \omega t)$. At the capacitor you have $U_{0C}=Q_0/C=i/(\mathrm{i} \omega C)$ since $Q=\int \mathrm{d} t i$ and thus $Q_0=i/(\mathrm{i} \omega)$. At the inductor you have $U_L=L \dot{i}$ and thus $U_{0L}=L \mathrm{i} \omega i_0$. Finally at the resistor $U_{0R}=R i_0$.

So you just calculate the circuit in terms of "complex resistances", $Z$, with the rule that $Z_R=R$, $Z_C=-\mathrm{i}/(\omega C)$, and $Z_{L}=\mathrm{i} \omega L$, i.e., you get
$$(Z_R+Z_C + Z_L)i_0=U_0 \; \Rightarrow \; i_0=\frac{U_0}{R-\mathrm{i}/(\omega C) +\mathrm{i} \omega L},$$
from which you can evaluate the amplitude
$$|i_0|=\frac{|U_0|}{\sqrt{R^2+(\omega L-1/(\omega C))^2}}$$
and the phase shift between $U$ and $i$ by
$$\phi=\text{sign} (\text{Im} i) \arccos \left (\frac{\text{Re} i}{|i|} \right).$$

 

[Frigus]

I don't know, what they mean by phasors. The usual way is to describe a circuit in the stationary state by complex currents and voltages (understanding that the physical signal is its real part).

Now take as an example the series circuit with a resistor $R$ and (ideal) inductor $L$, and a capacitor $C$. Now you take the external source as $U(t)=U_0 \exp(\mathrm{i} \omega t)$ and consider the quasistationary state, where also the current through the resistor, is of the form $i(t)=U_0 \exp(\mathrm{i} \omega t)$. At the capacitor you have $U_{0C}=Q_0/C=i/(\mathrm{i} \omega C)$ since $Q=\int \mathrm{d} t i$ and thus $Q_0=i/(\mathrm{i} \omega)$. At the inductor you have $U_L=L \dot{i}$ and thus $U_{0L}=L \mathrm{i} \omega i_0$. Finally at the resistor $U_{0R}=R i_0$.

So you just calculate the circuit in terms of "complex resistances", $Z$, with the rule that $Z_R=R$, $Z_C=-\mathrm{i}/(\omega C)$, and $Z_{L}=\mathrm{i} \omega L$, i.e., you get
$$(Z_R+Z_C + Z_L)i_0=U_0 \; \Rightarrow \; i_0=\frac{U_0}{R-\mathrm{i}/(\omega C) +\mathrm{i} \omega L},$$
from which you can evaluate the amplitude
$$|i_0|=\frac{|U_0|}{\sqrt{R^2+(\omega L-1/(\omega C))^2}}$$
and the phase shift between $U$ and $i$ by
$$\phi=\text{sign} (\text{Im} i) \arccos \left (\frac{\text{Re} i}{|i|} \right).$$

So what we are doing is that we are adding voltages across all components and then equating it to EMF of battery?
Now only thing I am not able to understand is the last step where the denominator is being square rooted.

 

[etotheipi]

Now only thing I am not able to understand is the last step where the denominator is being square rooted.

If $a, b \in \mathbb{C}$, then the complex number $z = a/b$ has a magnitude $|z| = |a| / |b|$. What is the magnitude of the complex number $R-\mathrm{i}/(\omega C) +\mathrm{i} \omega L$, and hence what is the magnitude of $U_0 / [R-\mathrm{i}/(\omega C) +\mathrm{i} \omega L]$?

 

[256bits]

I don't know, what they mean by phasors

Phasors are rotating vectors.
For the example, once one determines the 'phase' between the R, L, and C vectors, they all rotate in sync; a projection off to the right. ie axis-y gives its real value at a particular time t.

I think everyone has seen a vector rotating around the 360 degrees of a circle, with the projection forming a sine wave wrt to time t.

I forget why, maybe it is more visual than the more advanced math, which comes later I believe in the course.

 

[Charles Link]

One comment I have is that I'm a little surprised they are projecting the phasors onto the vertical axis instead of the horizontal axis to get the voltages. The way I learned it is the voltage is the real part of $\tilde{V} e^{ i \omega t}$, and the projection is along the horizontal axis.

That part of the text isn't shown in the OP, but they must be using $\mathcal{E}(t)=\mathcal{E}_o \sin( \omega t)$, instead of $\mathcal{E}(t)=\mathcal{E}_o \cos(\omega t)$.

 

[Frigus]

If $a, b \in \mathbb{C}$, then the complex number $z = a/b$ has a magnitude $|z| = |a| / |b|$. What is the magnitude of the complex number $R-\mathrm{i}/(\omega C) +\mathrm{i} \omega L$, and hence what is the magnitude of $U_0 / [R-\mathrm{i}/(\omega C) +\mathrm{i} \omega L]$?

Does this involves complex numbers?
I am very sure it is.
I haven't studied complex numbers, sorry.

That part of the text isn't shown in the OP, but they must be using E(t)=Eosin⁡(ωt), instead of E(t)=Eocos⁡(ωt).

They are using the same notation as you are using,right now I don't have access to the book as I am not at my home so I can be wrong.

 

[256bits]

One comment I have is that I'm a little surprised they are projecting the phasors onto the vertical axis instead of the horizontal axis to get the voltages. The way I learned it is the voltage is the real part of $\tilde{V} e^{ i \omega t}$, and the projection is along the horizontal axis.

That part of the text isn't shown in the OP, but they must be using $\mathcal{E}(t)=\mathcal{E}_o \sin( \omega t)$, instead of $\mathcal{E}(t)=\mathcal{E}_o \cos(\omega t)$.

Would it be that they picked the i-vector vertically - so when the R-phasor aligns with i-vector V_R = IR, which would be vertical axis projections.

 

[etotheipi]

Would it be that they picked the i-vector vertically - so when the R-phasor aligns with i-vector V_R = IR, which would be vertical axis projections.

I don't understand what you mean; you can't pick the "i-vector", it's just $i = e^{i\pi/2} \cong (0,1)$. Also, $V_R = IR$ holds for all time, not just when the $V_R$ phasor is vertical; the $I$ phasor also rotates with time.

 

[256bits]

I don't understand what you mean; you can't pick the "i-vector", it's just $i = e^{i\pi/2} \cong (0,1)$. Also, $V_R = IR$ holds for all time, not just when the $V_R$ phasor is vertical; the $I$ phasor also rotates with time.

The direction of the current axis is vertical is maybe better description.

 

[etotheipi]

The direction of the current axis is vertical is maybe better description.

What do you mean the current axis? There is no current axis, nor is there any voltage axis.

It's just a complex plane, onto which you may plot (time-dependent) complex numbers representing voltages, currents, whatever.

 

[hutchphd]

I believe this is why EE's use $j$ for the imaginary $\sqrt {-1}$.
Also they sometimes graph the (j)imaginary axis horizontally. So care must be taken in communication

 

[256bits]

It's been a while since working with phasors - I think it was j at the time.
My terminology is rusty.
I'll have to look back many years.

 

[vanhees71]

Phasors are rotating vectors.
For the example, once one determines the 'phase' between the R, L, and C vectors, they all rotate in sync; a projection off to the right. ie axis-y gives its real value at a particular time t.

I think everyone has seen a vector rotating around the 360 degrees of a circle, with the projection forming a sine wave wrt to time t.

I forget why, maybe it is more visual than the more advanced math, which comes later I believe in the course.

This was invented by some evil didactic guys to make the life of students hard ;-)).

Complex numbers are way simpler to deal with. It maps the analysis of AC circuits effectively such that you can treat them using Kirchhoff's rules known from DC circuits and just using the complex resistances (impedances) $Z$ for the usual elements (resistors, inductors, and capacitors).

 

[hutchphd]

My first introduction to AC circuits was phasors from Sears and Zemansky. It screwed up my understanding of complex impedance for quite a while. I do not understand why it is still taught but it is apparently in both Sears et al and Halliday et al. Can anyone justify it?

 

[Charles Link]

Phasors in ac circuit theory almost seems like magic, but is explained by linear response theory with $V_{out}(t)=\int\limits_{-\infty}^{t} m(t-t') V_{in}(t') \, dt'$, and the Fourier transforms satisfying $\tilde{V}_{out}(\omega)=\tilde{m}(\omega) \tilde{V}_{in}(\omega)$.

 

[vanhees71]

The only thing I hated in high school math and sciences were methods where one had to draw diagrams and read something off. The greatest nightmare was geometry, where you had to draw triangles and all kinds of constructions with "ruler and compass", and then I got some worse grade only because it wasn't accurate enough ;-)). In the final years, I had an agreement with my physics teacher that I was allowed to use complex numbers and dispensed from drawing these "Zeigerdiagramme", which you call "phasor".

 

[Charles Link]

To complicate matters, (this may look like extra information, but it might be useful for the OP in sorting things out if they read a couple of different textbooks on the topic), some EE texts will have $Z_L=-j \omega L$, and $Z_C=j/(\omega C)$. I spent a lot of time one weekend as a graduate student trying to make heads and tails out of this, and then I realized for this other sign convention their phasor diagram rotates clockwise instead of counterclockwise. In linear response theory, they essentially reverse the signs on the $e^{\pm j \omega t }$ of the Fourier transform and inverse Fourier transform. (The result is $-j \omega L$ points downward in the complex plane, but when rotating clockwise, it will lead the waveform for the electrical current by 90 degrees.)

Note again, most textbooks I believe use $\mathcal{E}(t)=\mathcal{E}_o \cos(\omega t)$. See post 15 above.

 

[vanhees71]

Yes, that's really confusing. I'm used to write $\exp(-\mathrm{i} \omega t)$ for harmonically time-dependent quantities (particularly in field theory and quantum theory that's the common convention). The only exception is for AC circuit theory, where at least the majority (if not all) German textbooks always use $\exp(\mathrm{i} \omega t)$. Of course, both conventions are completely equivalent, one must only be careful never to mix them within one calculation ;-).

 

[Charles Link]

To add to the above, in computing $\tilde{m}(\omega)$ for an RL or RC circuit, most textbooks use $\tilde{m}(\omega)=\int\limits_{-\infty}^{+ \infty} m(t) e^{-j \omega t} \, dt$, with inverse transform $m(t)=\frac{1}{2 \pi} \int\limits_{-\infty}^{+\infty} \tilde{m}(\omega) e^{+j \omega t} \, d \omega$, but reversing the signs in the $e^{\pm j \omega t}$ will result in $Z_L=-j \omega L$, etc. with a sign change for the complex impedances.

 

[vanhees71]

Shouldn't the upper limit be $+\infty$ (i.e., to get a Fourier integral), and sure, the same confusion using different conventions arises in the different use of the signs and factors in the Fourier transformation.

 

[vela]

So what we are doing is that we are adding voltages across all components and then equating it to EMF of battery?

Right. The basic problem is adding sines or cosines. If Kirchoff's voltage law gives you, for instance,
$$v = v_1 \sin(\omega t + \phi_1) + v_2 \sin(\omega t + \phi_2) + v_3 \sin(\omega t + \phi_3),$$ you can jump through hoops using various trig identities to evaluate the sum, but it's easier if we take advantage of Euler's formula and say $$v = \operatorname{Im}[v_1 e^{i(\omega t + \phi_1)} + v_2 e^{i(\omega t + \phi_2)} + v_3 e^{i(\omega t + \phi_3)}].$$ You can evaluate the sum using algebraic methods or you can represent each complex quantity in the complex plane and use what you know about adding vectors graphically. In general, either of these methods is less work than using trig identities. Once you have the sum, you just pick off the imaginary part and you have your answer. (If you were adding cosines, it's the same basic problem but you take the real part of the sum.)

Unlike the others, I think you should understand and know how to use phasors. It's not that difficult, and the more ways you know how to represent and work with complex numbers, the better. They can make some problems trivially easy. They do come in handy later on when you study wave optics, particularly when you get to single-slit diffraction.

 

[etotheipi]

@vela I don't think the issue is at all with phasors/complex numbers, it's more the strange formalism that this book and apparently others seems to use.

It just causes confusions, e.g. for example consider the title of this very thread: "why is the vector sum of peak voltages of circuit elements = peak EMF of the battery"

By labelling the phasors with their magnitudes, OP has likely been misled into thinking that a phasor represents a "peak" of that quantity, when in fact the word "peak" should be removed and the title should instead refer to the time-dependent voltages/voltage phasors themselves [and their time-dependent projections].

The "didactic" presentation in this book obfuscates what should be a very clear and simple treatment with complex numbers, and is ultimately unhelpful IMO.

 

[Charles Link]

To add to @etotheipi above, the OP @Hemant might even be missing the principal use of the ac circuit theory=Batteries are in general DC. The ac voltage refers to a voltage that is sinusoidal in time=the kind that is found in all electrical outlets.
(In general, the frequency used is normally $f=50$ or $60$ Hz, and the voltage amplitude is typically around 100 volts. )
The appliances and other things that the electrical outlet powers have impedances associated with them that are the result of resistive, inductive and capacitive circuit components. It may be worth mentioning that filament light bulbs are basically resistors. It can be good to have some introduction to ac circuits and their operation to at least get a basic understanding of how the electrical circuits work that we use on an everyday basis.

 

[etotheipi]

@Hemant, I feel that you might have been slightly left behind and that perhaps some of the discussion here and some of my earlier comments as well were not helpful for building your understanding. I'd like to try and summarise, just in case you're still a bit confused.

First consider an oscillating voltage $v = v_0 \cos{(\omega t + \phi_0)}$ applied to either a resistor, capacitor or inductor. Through the component a current will flow, oscillating each time at the same angular frequency but with perhaps an additional phase $\varphi$, that is $i = i_0 \cos{(\omega t + \phi_0 + \varphi)}$. Let's consider the AC response for resistors, capacitors and inductors in turn:$$\begin{align*}

\mathrm{R}: &\, i = \frac{v}{R} = \frac{v_0}{R} \cos{(\omega t + \phi_0)} \\ \\

\mathrm{C}: &\, i = \frac{dq}{dt} = C \frac{dv}{dt} = -\omega C v_0 \sin{(\omega t + \phi_0)} = \omega C v_0 \cos \left(\omega t + \phi_0 + \frac{\pi}{2} \right) \\ \\

\mathrm{I}: &\, i = i(0) + \int_0^{t} \frac{di}{d\xi} d\xi = i(0) + \int_0^t \frac{v_0}{L} \cos{(\omega \xi + \phi_0)} d\xi = \frac{v_0}{\omega L} \sin{(\omega t + \phi_0)} = \frac{v_0}{\omega L} \cos \left( \omega t + \phi_0 - \frac{\pi}{2} \right)
\end{align*}$$You may notice that the resistor response is in phase with the voltage, whilst the capacitor and inductor responses are $\pi / 2$ ahead and behind the voltage respectively.

It is convenient to introduce the complex voltage and complex current, defined by $V = V_0 e^{i \omega t}$ and $I = I_0 e^{i \omega t}$ respectively, where $V_0 = v_0 e^{i \phi_0}$ and $I_0 = i_0 e^{i (\phi_0 + \varphi)}$ are complex amplitudes encoding the magnitude and initial phase. The $\mathrm{Re}$ function maps from the complex voltage/current and standard voltage/current, that is, $v = \mathrm{Re}(V)$ and $i = \mathrm{Re}(I)$. One helpful property to remember is that the $\mathrm{Re}$ function commutes with time-differentiation; given some $z: \mathbb{R} \rightarrow \mathbb{C}$ with $t \mapsto z(t)$, you can write $z(t) = a(t) + ib(t)$ and thus$$\mathrm{Re} \left( \frac{\mathrm{d}z}{\mathrm{d}t} \right) = \mathrm{Re} \left( \frac{\mathrm{d}a}{\mathrm{d}t} + i \frac{\mathrm{d}b}{\mathrm{d}t} \right) = \frac{\mathrm{d}a}{\mathrm{d}t} = \frac{\mathrm{d}\left( \mathrm{Re}(z) \right)}{\mathrm{d}t} $$What this means is that you can do all the differentiations in complex numbers, and convert everything back only at the very end.

Now, consider a "black box" consisting of some network of resistors, inductors and capacitors. It may be assigned a (complex) impedance $Z$, which you may think of loosely as a complex analogue of resistance, and is defined as the ratio $Z := V / I$. For example, for the resistor you simply have $I = V/R$ and thus$$Z_R = \frac{V}{I} = R$$then for the capacitor, you have $I = C \frac{dV}{dt} = i\omega C V_0 e^{i \omega t} = i\omega C V$ and thus$$Z_C = \frac{V}{I} = \frac{1}{i\omega C }$$and for the inductor, you have $V = L \frac{dI}{dt} = i\omega L I_0 e^{i \omega t} = i\omega L I$ and thus$$Z_I = \frac{V}{I} = i \omega L$$How do you combine impedances? Let's just do one example, e.g. say $N$ impedances $\{Z_1, \dots, Z_N \}$ in parallel. The complex voltage $V$ across each of these is the same, and the sum of the complex currents through each must equal the total current into the parallel network, hence$$\sum_{k=1}^{N} I_k = V\sum_{k=1}^{N} \frac{1}{Z_k} = \frac{V}{Z_{\mathrm{eff}}}$$which upon division by $V$ yields the familiar formula.

You might wonder, do all of Kirchoff's laws, and the rules you know for analysing circuits hold also on the complex amplitudes $V_0$ and $I_0$? Yes! For example, consider the current law applied to a junction with $N$ inputs, $$(\forall t \in D) \quad \mathrm{Re}\left( \sum_{k=1}^N (I_0)_k e^{i \omega t} \right) \implies \sum_{k=1}^{N} (I_0)_k = 0$$because the only way for the sum to be zero for all values of $t$ is if the sum of the complex current amplitudes themselves vanish. The same analysis holds for the voltage law.

Why is this formalism useful? Well, let's consider your example of a circuit containing a resistor, capacitor and inductor in series, connected across an oscillating voltage source. You already know from the "impedances in series" principle that the resulting impedance is$$Z = Z_R + Z_L + Z_C = R + i \left(\omega L - \frac{1}{\omega C} \right)$$From this it's not hard to write down the current,$$I = \frac{V}{Z} = \frac{V}{R + i \left(\omega L - \frac{1}{\omega C} \right)} = \frac{V e^{- i \phi_Z}}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}}$$where the phase shift $\phi_Z$ is given by$$\phi_Z = \arctan \left( \frac{\omega L - \frac{1}{\omega C}}{R} \right)$$With this we can now work out some interesting quantities. For instance, what is the rate of energy dissipation in this oscillator, due to the resistor? You have$$\langle P \rangle = R|I_0|^2 \langle \cos^2{(\omega t - \phi_Z)} \rangle = \frac{1}{2}R |I_0|^2 = \frac{R|V_0|^2}{2(R^2 + (\omega L - \frac{1}{\omega C})^2)}$$At resonance, $\omega_0 L - \frac{1}{\omega_0 C} = 0$. Half-power thus occurs when $\omega_h L - \frac{1}{\omega_h C} = \pm R$, or in other words when$$\omega_h = \mp \Gamma + \sqrt{\Gamma^2 + \omega_0^2}$$with $\Gamma := R/2L$. Then, you may identify the full-width half-power as $\delta \omega = \frac{R}{L}$, which results in a quality factor $Q = \frac{\omega_0}{\delta \omega} = \frac{1}{R} \sqrt{\frac{L}{C}}$.

In other words, complex numbers allow you to fairly easily analyse key properties of oscillating circuits, and oscillating systems in general.

 

[Charles Link]

One correction to post 34=the electrical current is always the same in each component of the series circuit. It is the voltages that are out of phase with the current for the inductor and capacitor. For the resistor, the voltage drop across the resistor is in phase with the current. The idea is good, but it might need a minor edit or two. (Edit: See post 36=very good).