# Tag Info

20

The identity $$V = K \frac{dV}{dt}$$ is only guaranteed with a constant $K$ if your assumptions actually hold. The first identity $V=RI$ only holds for a resistor, while the other holds for a capacitor. So in this sense, the letters $V,I$ in these equations mean something else. In one of them, it's the current through (or voltage on) a particular resistor, ...

14

This gives me a feeling that a capacitor never gets charged fully. Am I right? Why not? In the context of ideal circuit theory, it is true that the current through the capacitor asymptotically approaches zero and thus, the capacitor asymptotically approaches full charge. But this is of no practical interest since this is just an elementary ...

13

Electrical analogies of mechanical elements such as springs, masses, and dash pots provide the answer. The "deep" connection is simply that the differential equations have the same form. In electric circuit theory, the across variable is voltage while the through variable is current. The analogous quantities in mechanics are force and velocity. Note that ...

12

The maximum charge a capacitor stores depends on the voltage $V_0$ you've used to charge it according to the formula: $$Q_0=CV_0$$ However, a real capacitor will only work for voltages up to the breakdown voltage of the dielectric medium in the capacitor. So in reality, for every capacitor there is a maximum possible charge $Q_{max}$ given by: $$... 10 Short answer: this is a textbook example of the limitations of ideal circuit theory. There seems to be a paradox until the underlying premises are examined closely. The fact is that, if we assume ideal capacitors and ideal superconductors, i.e., ideal short circuits, there appears to be unexplained missing energy. What's not being considered is the ... 10 If you redraw your diagram as: It should be clear which capacitors are in parallel and which are in series. 9 You can calculate the energy by assuming you charge the capacitor with a constant current. Then the energy input to the capacitor is \int_0^t{IV(t')\mathrm{d}t'} Since I is constant, you know that (for a linear capacitor) V(t) is a ramp function. So from geometry (the area of a triangle formula) the integral is \frac{1}{2}ItV(t) Now, It is the ... 9 A battery generates a voltage by a chemical reaction. There is a class of chemical reactions called redox reactions that involve the transport of electrons, and you can use the reaction to drive electrons through an external circuit. This is the basis of a battery. The battery will continue to provide power until all the reagents have been used up and the ... 8 The three capacitors are connected in parallel. There are only two nodes in this circuit. A series connection requires at least three. The equivalent capacitance is just the sum of the three capacitances. UPDATE: The circuit can be redrawn such that the parallel connection is manifest. 8 If you have just given the voltage signal with$$ \def\l{\left}\def\r{\right} v(t) = \l(2-\l|\frac t{\rm s}-2\r|\r)\rm V $$then the current at t=2\rm s is undefined. Right. But, in most cases really nobody cares. What we learn theoretically about the current from the above voltage signal definition is that$$ i(t) = \begin{cases} C\cdot 1\frac{\rm V}{\rm ...

7

This is true but strictly limited to RC circuits without external sources: that is, a resistor hooked up to a capacitor with nothing else in between. In that case, $V$ is indeed proportional to $\dot V$, with a crucial minus sign in between: $$\dot V=-\frac1\tau V,$$ where $\tau>0$ is some constant. This equation implies that $V(t)=V(0)e^{-t/\tau}$, ...

7

The physical meaning of the capacitance is precisely given by $\mathrm{d}Q=C\cdot \mathrm{d}V$: $C$ tells you how much charge there will be in the capacitor per voltage applied. For all capacitors, the linearity holds fairly well. Generally speaking, capacitance is given by a Q-V curve, which may consist of a linear region, a saturation region and a ...

7

First, note that the light bulb is essentially just a glorified resistor. As current flows through the filament, Joule heating causes the filament to get hot and emit light. When one places a capacitor in a circuit containing a light bulb and a battery, the capacitor will initially charge up, and as this charging up is happening, there will be a nonzero ...

6

Everything you've probably learned about capacitors, especially including the statement that opposite plates of the capacitor carry opposite charges, applies only to a capacitor in a circuit. If your capacitor is floating, so that the plates are not connected to anything, the charge on the plates is not going to change. If you hook up only one plate to a ...

6

Power lines do cause corona discharges (power line inspection video), which produce some UV light. If the camera is picking up some UV light that might cause a few dots on the picture. If you inspect the pictures closer you will notice that the pattern only appears sometimes and only in two of the cameras of the car (the front and the rear facing camera). ...

6

I don't think these distortions are necessarily caused by the power lines. Going along some way forward on your link, under the power lines and then some more, you get to this image, which shows the distortion in front of the car over an area far from the power lines, when the car is also pretty far from them. EDIT: There's also some distortion inside ...

6

Generally, no. You've written two equations. The first relates voltage and current for an isolated resistor. The second relates voltage and current for an isolated capacitor. Given only those two expressions, there's no reason at all that they can or should be combined. That is, you have no circuit, only isolated components. There are principles for ...

6

Capacitors, as used in electric circuits, do not store electric charge. When we say a capacitor is charged, we mean energy is stored in the capacitor and, in fact, energy storage is one application of capacitors. Now, for an ideal capacitor in a circuit context, the current through is proportional to the rate of change of the voltage across: $$i_C = C ... 5 Does that mean when I apply a voltage, the current will be infinite large? No, not even in the context of ideal circuit theory. It's a bit subtle since we're using phasor voltages and currents and that requires a couple of assumptions to hold in order to be valid. When those assumptions don't hold, we have to see what the 'infinity' (division by ... 5 When the switch is opened, the circuit is the equivalent of this, so I think you can clearly see that the resistors are in series and so are the capacitors. It seems that you already understand how to calculate series resistance, so I'll show how to understand series capacitance. First, you probably already know that capacitance is defined as the ratio ... 5 The only property of metals used in deriving C=\varepsilon A/d is that they are perfect conductors. Ideally, all metals have this property. So even if you change the metal, it should not matter. But if you use something other than metal, then it will of course change the capacitance. 5 Suppose you have two conductors kept at voltages V_1 and V_2 and they have charges Q_1 and Q_2 respectively. Then, one has the relation$$ Q_1 = C_{11}\ V_1 + C_{12}\ V_2\quad \textrm{and}\quad Q_2 = C_{12}\ V_1 + C_{22} \ V_2 \ ,  which defines a (symmetric) Capacitance matrix that is determined by the geometries of the two conductors. This ...

5

Here is my understanding: When you increase the distance between electrodes - capacitance drops, but stored charge remains the same, as electrons have nowhere to go. Same charge in lower capacitance means higher voltage potential. Without that part of stored energy would just vanish :-)

5

Toby, I agree that this is really counter intuitive and I was also quite surprised as well when I first saw this very demonstration. I am an undergraduate TA and this is how I explained it in my lab section. I hope this helps. I see two parts to a full explanation: (1) Why is the electric field constant and (2) why does the potential difference (or voltage) ...

5

There is no such thing as inherent capacitance of a object. Capacitance is a property between two conductors. Your question therefore makes no sense. Capacitance of a typical human to where? In the case of touch screens, the sensors are measuring capacitance back to their own ground. How well a human touching a sensor couples back to that ground has a ...

5

As the two charged bodies attract, they have unlike charges. So, Assuming your two charged bodies as conductors and charged equally, the system may be considered as a Capacitor. If you place a dielectric like glass of some Relative permittivity $\epsilon_r$ (3.7 to 10) which fills the empty space between the bodies, then the capacitance would be ...

5

Crazy Buddy's answer and related comments have made the point that you could indeed use a capacitor to charge a battery, but the amount of energy stored in capacitors is generally less than in batteries so it wouldn't charge the battery very much. However there is a new generation of capacitors called ultracapacitors that are being developed with electric ...

5

In general, no. Specific gravity refers to the density of a substance - how heavy a liter of it is. Dielectric constant refers to the response of the substance to an electric field, which depends on the chemistry of the substance. They are not physically related in any obvious way. It is possible to find two liquids with very similar densities and ...

5

Every system likes to decrease its electrostatic energy. The charges on the plates are almost in stable equilibrium. The charges on the opposite plates attract them, and the charges on the same plate repel them with almost the same force. However, a capacitor has fringe fields: These may be negligible when calculating the field inside a capacitor, but ...

5

First of all, thanks for this question because it made me think about relativity which was always fun! It's true that $E'=\frac{1}{\gamma} E$. You say that relativity states that the energy should increase by a factor of $\gamma$. This is certainly true for a massive particle whose energy is $\gamma mc^2$, but why would you expect this to hold for the ...

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