In AC, there are $R$, $C$, &amp; $L$ or $R$, $X_C$, &amp; $X_L$. $$ X_C = \frac{1}{\omega C} $$ where $\omega=2\pi f$.  But sometime we're using the $X_C = \frac{1}{j\omega C}$. I understand the $j$ is the imaginary unit and it's current lead voltage in capacitance circuit. But why does the equation sometimes have $j$ and sometimes don't?
99319
As you know, impedance is defined as a complex number.  Ideal capacitors: $$ \frac {1} {j \omega C} \hspace{0.5 pc} \mathrm{or} \hspace{0.5 pc} \frac {1} {sC} $$  Ideal inductors: $$ j \omega L \hspace{0.5 pc} \mathrm{or} \hspace{0.5 pc} sL $$  I know that the reason why they 'invented' the concept of impedance is because it makes it easy to work with circuits in the frequency domain (or complex frequency domain).  However, since in real-life circuits both voltages and currents are real numbers, I'm wondering if there is any actual physical meaning behind the imaginary component of impedance.
122684
I'm studying the following circuit   The image is explicit but as you can see the input measured in the upper terminal is a square wave with \$E_c\$ and \$-E_c\$ as levels.   Now I also had no trouble understanding that if \$R_B \leq \beta R_C\$ when the square wave is in the positive half-cycle we have saturation and when it is on the negative half-cycle we have cutoff situation.  Now in saturation we will (because the junction voltages are zero) an output voltage of zero. That is correct with the figure. However when we are in cutoff, applying KVL to the output mesh leads us to   $$U_{CE}=-E_c$$  because currents can be neglected.  That however is not concordant with the figure which shows $$U_{CE}=E_c$$ Why is that? What am I missing?
36150
If we write out Maxwell's equations with magnetic charges, we get $$ \begin{align} \nabla \cdot \mathbf{E} &amp;= 4 \pi \rho_e \tag{1}\\ \nabla \cdot \mathbf{B} &amp;= 4 \pi \rho_m \tag{2}\\ -\nabla \times \mathbf{E} &amp;= \frac{\partial \mathbf{B}}{\partial t} + 4 \pi \mathbf{J}_m \tag{3}\label{Eq:Faraday}\\ \nabla \times \mathbf{B} &amp;= \frac{\partial \mathbf{E}}{\partial t} + 4 \pi \mathbf{J}_e \tag{4}\label{Eq:Ampere} \end{align} $$ In particular, Faraday's law \eqref{Eq:Faraday} contains a minus sign that Ampere's law \eqref{Eq:Ampere} does not. This always struck me as odd because it's often said the fields are dual to each other (i.e. you can replace E with B and &quot;get the same result&quot;), but that requires a bit of mental recalibration to accommodate that minus. So I'm curious what the origin of that negative is and what it means. Are there any intuitive explanations for how to think about it?
82109
I am trying to drive  with an  microcontroller.  Both devices are listed to work with I2C clock frequencies up to 400 kHz. The two devices are the only devices on the I2C bus.  However, working out the calculations for the pull-up resistor bounds gives some rather odd values.  Calculating the I2C minimum pullup resistor value:  \begin{equation} R_{min} = \frac{Vcc - 0.4}{3mA} = 966.7 \Omega \end{equation}  Looking at the uC datasheets, on page 92 lists the max pin input capacitance of 10pF.  However for the LCD, it has on page 8 something called the Capacitive load represent by each bus line labeled as Cb, and is listed at a max value of 400pF. I'm assuming I should just add this value to the 10pF uC input capacitance, but this seems really high and the calculations are wonky.  For example, when I try computing the maximum pullup resistor value for a 400kHz clock:  \begin{equation} R_{max} = \frac{300ns}{10pF + 400pF} = 731.7 \Omega \end{equation}  Am I misinterpreting the LCD datasheet? Obviously the max allowable pullup resistor value cannot be smaller than the min allowable value.  likewise, if I assume a maximum net bus capacitance of 400pF, I get:  \begin{equation} R_{max} = \frac{300ns}{400pF} = 750 \Omega \end{equation}  still under the maximum allowable value.
38356
I just started learning about electronics and decided to quickly sketch a circuit. When it comes to OP and changing the said circuit into an Ac equivalent, I got stuck. Sorry in advance if this is considered a basic question.
69427
I need to find node voltage at A. This seems like a trivial question, but when I set up the resistive voltage divider equations they are not equal to each other. Am I doing something wrong?  My equations are \$V_A = V_1(R3/(R3+R1)) = V_2(R3/(R3+R2))\$      I tried solving this problem with an LTSpice simulation and with the superposition method, but my answers did not match (although very close to each other). Is this because the simulator uses an approximation? Can someone compare results?
95566
What is the potential diff between two plates of charges $q_1$ and $q_2$ ($q_1$ $&gt;$ $q_2$)? I know the equation is $V = \frac{q_1-q_2}{2C}$ where $C$ is the capacitance of the system but how has this equation been derived?   I apologize if this is a duplicate question. Google searches with different search terms didn't yield any answers.
121569
From this paper:     Shouldn't $du$ be $dt$? And also how do you get from that line to the final result if $du$ is not $dt$?
112120
I want to find the total resistance between A and B. This is a HOTS mcq, which has the following options:   A: 45\$\Omega\$ B: 15\$\Omega\$ C: 5\$\Omega\$ D: 22.5\$\Omega\$   A book says that the answer is 5\$\Omega\$ as R1, R2 and R3 are parallel. But I cannot understand how can they be parallel? Our science teacher said that R1, R2 and R3 are in a series, so he said that total resistance R=45\$\Omega\$. The book that says that the answer is 5\$\Omega\$, does not have the method to find it out.  I tried that:   All are in series :- 45\$\Omega\$ 15\$\Omega\$ is parallel with 15+15+15\$\Omega\$ and with one more 15\$\Omega\$, which would turn into : 45/7\$\Omega\$, which is not possible. R3 is parallel with R1 and R2, as well as in a series with R2. R1 is parallel with R2 and R3, as well as in a series with R2 and R3. But then I cannot determine the method to get the answer.   So I am confused between 5\$\Omega\$ and 45\$\Omega\$. I think as the book says, the answer must be 5\$\Omega\$, but what is the method?
114598
What is actually alternating current (AC)? And direct current (DC)? I've been reading several books about basic electric circuits analysis and they are contradictory.   Some books say that AC is the same as the sinusoidal stationary state. That is, a sine or cosine function (actually, they are the same only shifted horizontally). For instance, \$v(t) = A \sin(\omega t + \phi)\$. Acording to these books, DC is totally stationary (static), that is, the voltage and current are not time-dependant (\$v(t) = k\$, where \$k = \mathrm{const.}\$). Currents have the same form as voltages on each of these cases. According to other books, AC is any current that is changing (that is alternating) his sign over time (\$t\$). With maths notation: \$i(t_1) &lt; 0\$ and \$i(t_2) &gt; 0\$ for some \$t_1\$ and \$t_2\$ where \$t_1 \neq t_2\$. It could have any shape as long as it's alternating his sign.   I'm inclined to think that in practice wins the first option but maybe it's better the etymology of the second option.  Some other cuestions have relation to this:   Is AC a voltage that is the sum of a sinusoidal stationary state and a constant voltage? For instance, \$v(t) = A \sin(\omega t + \phi) + k\$. Is AC a shawtooth wave?
89751
The usual Lorentz force expression I am familiar with is this:  $$\vec F=q(\vec E+\vec v \times \vec B)$$  I have seen some other versions lately that include an extra factor $1/c$:  $$\vec F=q\left(\vec E+ \frac{1}{c} \vec v \times \vec B\right)$$  What is this $c$ and how is it included? I guess other parameters in the expression are also different from the top expression for this to fit?  Example of a text snippet where I have run across this extra parameter:
113414
In $\triangle ABC$ with $AB=AC$ and $\angle BAC=20^\circ$, point $D$ is on $AC$, with $BC=AD$. Find $\angle DBC$.   I know the correct solution, but I'm more interested in where is the problem in my solution.  My solution :    Now in $\triangle ABD$, applying the sine rule:  $$\frac{AD}{\sin\alpha} = \frac{BD}{\sin 20^\circ} \tag{1}$$  In $\triangle BDC$:  $$\frac{BD}{\sin 80^\circ} = \frac{BC}{\sin(180^\circ-\beta)} \tag{2}$$  We know $AD= BC$; put in to $(1)$:  $$\frac{BC}{\sin\alpha} = \frac{BD}{\sin 20^\circ}  \tag{3}$$  Comparing $(2)$ and $(3)$:  $$\frac{BC}{BD} = \frac{\sin\alpha}{\sin 20^\circ} = \frac{\sin(180^\circ-\beta)}{\sin 80^\circ} \tag{4}$$  $$\frac{\sin \alpha}{\sin(180^\circ-\beta)} = \frac{\sin 20^\circ}{\sin 80^\circ} \tag{5}$$     Now, $\alpha = 20^\circ$ and $\beta = 100^\circ$, but when I plug these values in $\triangle ABC$, it's not even triangle. oO   Where I am wrong? Thanks.  PS : sorry for poor editing, I don't have any clue about it.
122620
In an alternating current, the flow of electric charge periodically reverses direction, and the number of times it does that is called the frequency of the current. However, if the frequency of an AC is made infinite, would any current flow? I asked this question to my school teacher and he was unable to provide me with an answer unfortunately.  If I consider a horizontal wire, through which an AC is flowing, then I can say that an electron in the wire is moving 50 times left and 50 times right, if the frequency is assumed to be 50Hz. It seems to me that if I increase the frequency, the displacement of the electrons in both the directions, that is left and right, will gradually decrease, and finally become 0, when frequency is infinite, thus making I = 0.  What have I got wrong?
79830
Suppose a series RLC circuit has its resonance frequency f_r. If we apply AC current of frequency f_r to this series RLC circuit then circuit has maximum current as output. My question is little bit hypothetical. What will be the output of the series RLC circuit if we apply AC current of mixed frequency where one of its frequency is f_r? For simplicity we can assume AC current made up of two frequencies, (say f_r and f_s) where f_r != f_s. PS: I am very sorry for any incovenience. Please comment if still there is ambiguity.
75511
In context to the diagram below , it is quite easy to find the current flowing through section AD , which turns out to be 2A , however , I cannot determine the current in any other branches with Kirchoff's Voltage Law . The branches AB,CD,DE and FA have no resistance across them so whenever I take those branches in any of the the possible loops , the potential drop/rise will always be 0 so the terms involving 'I' never appear and consequently they are indeterminable.So how do I find the magnitude of 'I' ? In this particular case I can guess than I=1A with junction rule and the symmetry of the circuit - the batteries have same emf, so what if the emfs aren't same ? then the original problem comes up again.
121910
The formula for resistivity is: $$\rho = \frac{RA}{L}$$  where $\rho$ is resistivity,  $R$ is resistance, $A$ is cross-sectional area, and $L$ is the length of the conductor.    We can see from the formula that $A$ and $L$ are involved, why then does resistivity not depend on dimensions?
7843
Assume a test charge with charge $Q$ is placed at the center of a 13-sided regular polygon, the vertices of which are charges with equal charge $q$. What is the electrical force on the center charge?  To start we write this as our electrical force... $$ \vec{F}_{Q} = \sum^{13}_{n} \frac{1}{4\pi\varepsilon_{0}}\frac{qQ}{R^{2}} \left(cos\left(\frac{2n\pi}{13}\right)\hat{i} , sin\left(\frac{2n\pi}{13}\right)\hat{j}\right)  $$   and with some simplification we get... $$ \vec{F}_{Q} = \frac{1}{4\pi\varepsilon_{0}}\frac{qQ}{R^{2}} \left\{1-2cos\left(\frac{\pi}{13}\right) + 2cos\left(\frac{2\pi}{13}\right)-2cos\left(\frac{3\pi}{13}\right) + 2sin\left(\frac{\pi}{26}\right)-2sin\left(\frac{3\pi}{26}\right)+2sin\left(\frac{5\pi}{26}\right), 0\right\}$$ ...more simplification... $$ \vec{F}_{Q} = \frac{1}{4\pi\varepsilon_{0}}\frac{qQ}{R^{2}}  \left\{ 1.11022\times10^{-16}, 0 \right\} $$ Now we have an electrical force on the test charge, however, in a ring of equidistant charges such as the one mentioned above, shouldn't the $\hat{i}$ component be equal to zero? Does anyone have any ideas as to why I'm getting a non-zero number for my $\hat{i}$ component?