In a circuit diagram were capacitance value of an electrolytic capacitor is given, how do you know the capacitor voltage to use?
17221
How to find the voltage rating of a capacitor after knowing the capacitance value.  Example, For a power supply, If the input voltage is 60V and output is 5V. The input capacitor(C1) value is 100uf and output capacitor value is 1000uf.  What is the voltage rating for both the capacitors and how to find them.
46052
I understand that the output voltage is determined by the ratio between the two resistor values, and that if the both resistors are same then output voltage will be exactly same for all; but what's the basis of picking the resistor values?there is any need of consider output current to choose resistor value.
91099
When a capacitor is connected to battery, it stores $\frac{C V^2}{2}$, while battery supplied $CV^2$ energy. Therefore, $\frac{C V^2}{2}$ energy gets lost as heat. When a capacitor is already charged and a dielectric is inserted in this charged capacitor (which is still connected to the battery), will there be any heat produced ?
6148
Most capacitors are in the µF, nF and pF range. I know there are some special ones that go that high, but at the time faraday was around, and the unit was named after him, they didn't have such a thing. Why is the unit so large if we rarely use caps with that high of a value?
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.
87401
If you imagine a dielectric of height $d$ and dielectric constant $\kappa$  placed evenly in a capacitor of height $D$. We define $\Delta x = \frac{D-d}{2}$.   The equivalent capacitance is now  $$\begin{align} \frac{1}{C_{eq}} &amp;= 2 \frac{\Delta x }{\epsilon_0 A} + \frac{d}{\kappa \epsilon_0 A} \\ &amp;= \frac{D-d}{\epsilon_0 A} + \frac{d}{\kappa \epsilon_0 A} \\ &amp;= \frac{\kappa(D-d)+d}{\kappa\epsilon_0 A} \end{align} $$  At this point   $$C_{eq} = \frac{\kappa\epsilon_0 A}{\kappa(D-d)+d}$$  Based on prior knowledge, it should be that the charge on the equivalent capacitor is the same as the charge on the original, non dielectric capacitor  In other words:  $q = CV = C_{eq}V$ and voltage hasn't changed.   But $CV \ne C_{eq}V$!  What am I missing?
64319
I have a 9V AC source which will have a max load of 100mA.  I'm trying to figure the value of a smoothing capacitor with less than a 10% ripple.  I don't have a scope or else I'd just try a bunch out.  I'm just a hobbyist, but I'm trying to understand this.  Since I'm not sure if I'm doing this right, I'll just step through what I've done so far.  First I found this formula: $$V_{ripple}={V_p \over R_LC} \times \Delta T$$  I'm using 60Hz 9VAC that is full wave rectified and I want a 10% ripple so:  $$(9V \times .1)={9V \over R_LC} \times {1 \over 60Hz}$$  I'm calculating my resistive load with Ohm's law. $$R = {9V \over .1A} \quad R = 90 \Omega$$  So plugging that in I get: $$.9V={9V \over 90\Omega \; C} \times {1 \over 60Hz}$$  If I solve for C (admittedly using wolfram alpha), I get: $$C={1\over 540}F \quad C=1851 \mu f$$  1850µf seems pretty high to me...  And if I lower the ripple percent it get crazy high.  Is this the right way to find this?  Or did I screw up somewhere?  Also, my circuit has the capacitor between the full bridge rectifier's positive and negative outputs.  I've only seen this type of smoothing circuit, but maybe my circuit needs to change?
9441
The maximum energy stored by a capacitor is when it’s (theoretically) fully charged, or just before it reaches dielectric breakdown, if you will. It seems, then, that the maximum energy ($E$) $=QV$, where $Q$ is the maximum charge attained and $V$ is the maximum ‘voltage’. However, the correct equation for the maximum energy is $E=\frac{1}{2}QV$. Why? I’m looking for a simple, intuitive explanation because my reasoning that $E=QV$ seems sound to me. I know about , but I’m looking for a reason to disprove my specific argument that $E=QV$ and not the “Oh, it comes from integration.” response.
76144
If you connected the positive terminal of a battery to the negative terminal to a battery with a wire with (hypothetically) no resistance, and are asked to give the voltage drop of a segment of wire in the circuit, how would you determine this?
123028
What exactly would happen in the ideal case where a capacitor were to discharge into a wire with zero resistance?  Like in here:  There is still charge on one plate of the capacitor, which will try to go to the other plate to make $\Delta V = 0$ across the capacitor, so there has to be a current $I$.  But as more charge goes to the other side of the capacitor, the more repulsion these will cause to more coming towards them. So I am assuming current cannot just be constant?  I am starting from $$ I = C \frac{dV}{dt}, $$ but I know neither $V(t)$ nor $I$.
1104
I'm trying to work out what value resistor to use in a LED circuit. The equation I'd use to do this is:  $$ R = \frac{V_{cc} - V_f}{I_f} $$  Seems logical, and makes complete sense. The answers to the question  confirm this too.  I have the following LEDs:   \$ V_f = 3.3V \$ \$ I_{f_{typ}} = 20mA \$   Using a 5V power supply:   \$ V_{cc} = 5V \$   Plugging these into the above equation gives:  $$ \begin{eqnarray} R &amp; = &amp; \frac{V_{cc} - V_f}{I_f} \\   &amp; = &amp; \frac{5V - 3.3V}{20mA} \\   &amp; = &amp; 85\Omega \end{eqnarray} $$  All good so far.  However, if I use the calculator at , that gives me 100&Omega;. If I use the ElectroDroid app on my phone, that gives me 85&Omega;.  So, I assume that the linear1 calculator is using a different method of calculating this resistor value; is there some better way of doing this?
56120
Hey guys I cannot wrap my head around this damn circuit and how I'm supposed to solve it using nodal analysis. I chose the bottom node as my reference node and I got to these equations:  Node A is the left one, node B is the middle one, node C is the right one and the reference node is the bottom one.  Node A :        (Va/100)  -4  + (Va-Vb)/50 + (Va-Vc)/20 = 0  Node B :        -(Va - Vb)/50 + 4 - 10 -2 + (Vb-Vc)/40 = 0  Node C:         2 + (Vc/25) -(Vb-Vc)/40 -(Va-Vc)/20 = 0  I am supposed to find Vx.  Are my equations right? How do I find Vx?
99225
I'm having difficulty understanding why a floating capacitor will store less charge than a grounded capacitor. Imagine you have two parallel plates and a low DC voltage source like 5V, with the negative side connected to neutral ground, and that you have two different ways of wiring it: In case 1, you connect one side of the voltage source to each plate to make a normal capacitor. The voltage across the capacitor is given by $Q = C V  $ is given by the geometry of the plates. In case 2, you still connect one plate to +5V, but leave the other plate floating and neutral. According to my understanding, charge should still build up on the positive side of the capacitor because it's at the same potential relative to the other plate as case 1, and it has the same geometry. On the negative plate, negative charge should be locally attracted to the positive plate, leaving behind positives, overall keeping the plate neutral. Practically, I'm pretty sure case 2 is wrong. What I think will happen is a miniscule amount of surface charge will be stored in the positive wire to maintain the electric field, but that charge will be way less than the stored charge of case 1. Can you explain where my understanding is wrong/not deep enough?
41243
There're some easy concepts which I don't get exactly in my mind. I'm afraid I've been studying these things for two years of my engineering but they still bother me. Capacitor is one of them. Can someone explain?    What does a capacitor do? Does it store charges? If so, then how does it do?   I have searched it on Google and Yahoo but didn't find any helpful thing there (for me). So I will be glad if I got my problem solved here.  P.S. I hope that the question would not again be an off-topic, as it always does and moreover people don't suggest then where to go. It's a real sad thing.
26562
I'm a beginner in electronics from the software industry. With some self-taught things, I'm trying to implement some basic Arduino circuits. My confusion is about capacitors mainly. My understanding about capacitors is that they act as power storage for few seconds or milliseconds.  I found that most IC's must have capacitors connected to their pins.  My confusion is how to find out which pins need a capacitor, and how to find the correct capacitor for a circuit or a capacitor for an IC.  Finally, why are capacitors necessary in a circuit in such situations?
158
For a simple circuit with a battery supplying a voltage $V$ to a capacitor, let us assume that the charge on the capacitor is $Q$. Now, the work done by the battery or the energy supplied is given by the relation:  $$W=QV$$  But the energy stored in the capacitor is given by:  $$U = \tfrac12 QV$$  The value of $Q$ as well as that of $V$ should be the same in both the equations.  Now my question is, where is the other half of the energy that the battery supplied?
42598
There was some discussion on this question        which I don't see as being conclusively resolved:   "turns out that what might LOOK like two ordinary electrolytics are not, in fact, two ordinary electrolytics." "No, do not do this. It will act as a capacitor also, but once you pass a few volts it will blow out the insulator." 'Kind of like "you can't make a BJT from two diodes"' "it is a process that a tinkerer cannot do"   So is a non-polar (NP) electrolytic cap electrically identical to two electrolytic caps in reverse series, or not?  Does it not survive the same voltages?  What happens to the reverse-biased cap when a large voltage is placed across the combination?  Are there practical limitations other than physical size?  Does it matter which polarity is on the outside?  I don't see what the difference is, but a lot of people seem to think there is one.  Summary:   As posted in one of the comments, there's a sort of electrochemical diode going on:     The film is permeable to free electrons but substantially impermeable to ions, provided the temperature of the cell is not high. When the metal underlying the film is at a negative potential, free electrons are available in this electrode and the current flows through the film of the cell. With the polarity reversed, the electrolyte is subjected to the negative potential, but as there are only ions and no free electrons in the electrolyte the current is blocked. —    Normally a capacitor cannot be reverse-biased for long, or large currents will flow and "destroy the center layer of dielectric material via electrochemical reduction":     An electrolytic can withstand a reverse bias for a short period, but will conduct significant current and not act as a very good capacitor. —    However, when you have two back-to-back, the forward-biased capacitor prevents a prolonged DC current from flowing.  :     For circuit positions when reverse voltage excursions are unavoidable,   two similar capacitors in series connected “back to back” ... will create a   non-polar capacitor function ... This works because almost all the circuit voltage   is dropped across the forward biased capacitor, so that the reverse   biased device sees only a negligible voltage.   :     The oxide dielectric construction that is used in tantalum capacitors   has a basic rectified property which blocks current flow in one   direction and at the same time offers a low resistance path in the   opposite direction.