# Tag Info

1

If some charge is given to a conductor then its potential will be remain same through out the region, because work done on every charge is same.

3

Why do we use capacitors when batteries can very well store charges? There's an important point that, so far, I don't see in other answers. Neither of these devices store charge! A "discharged" battery or capacitor contain the same net quantity of electrical charge as a "fully charged" battery or capacitor. What they are "charged" with is energy, not ...

2

A capacitor stores charge on a pair of plates. A battery generates charge through chemical reactions that break neutral atoms into positive and negative ions. Both store energy. A battery stores chemical energy. A capacitor stores potential energy in the separated charges. Sometimes a capacitor has an electrolyte between the plates. This is a molecule ...

2

batteries are a much more efficient at storing electricity but in circuits, it makes much more sense to use capacitors in circuits as they are much more efficient for the short term storage of electricity. batteries are a lot more bulky and to work as a capacitor they would need to be rechargeable. it would not make sense to have two batteries in a single ...

4

While a capacitor can be used to store charge, usually we are interested in other properties. Most notably, it has a voltage proportional to the amount of charge stored ($Q=CV$) which means it acts as an integrator of current. There are many circuit applications where you use this property - which incidentally also means that the apparent impedance of a ...

3

Practically we use capacitors when we require a large amount of charge to be flown within fractions of seconds.. Battery provides a nearly uniform voltage and effective in long use, but when it comes to discharge a large amount of charge in a fraction of second, battery is ineffective.. How ever by a building a capacitor with large capacitance we store a ...

2

I've to make an electronic circuit 'RC' and the relation between current and tension between two nodes must be fulfilled by a capacitor 'C' (it integrates the current; see the relation in the WP). I can use a battery with a constant tension to power the circuit ($V_{in}$in the second image) but not to model that relation.

1

note: I accidentally thought OP was asking about a series $LC$, not a series $LCR$. Including the $R$ changes the results here by making the infinities turn into large finite values. Suppose you hook your series $LC$ circuit up to a voltage source with frequency dependent phasor $\tilde{V}_s(\omega)$. Intuition First let's guess what happens. At low ...

0

Say your series RLC circuit is excited by a constant current source producing current $I$ at angular frequency $\omega$. Since all 4 elements (R, L, C, and source) are in series, the current through any one of them is just the source current. Then for each element, $V_n=IZ_n$. For an inductor, the impedance Z is given by $Z=i\omega{}L$. For a capacitor, ...

0

Please look here you can make a function of w for both voltage across inductor and capacitor, and you can check the neighbourhood of resonant freq.

1

This would actually be easier to answer over at the EE stackexchange site since there is a handy schematic editor built in. First, note that, by speaker wire, we're actually referring to a speaker cable; in this case, a pair of wires. For each wire, we can assign a series resistance and inductance (per foot), i.e., the $R$ and $L$ of each wire is in series ...

0

When using Gauss's law to calculate the flux through a closed surface we take the field component in the normal direction of the surface. The normal direction always points outward for the closed surface. Going by the math, it seems like you're making a Gaussian cylinder that encloses one of the plates. The difference between making a cylinder that ...

1

In computer science, there's a semi-joking saying that $\log n < 50$ for all $n$. Of course this isn't true – as you say the logarithm is unbounded. But what is true is that it grows so slowly that, if you can only put in quantities like memory size, mass of some material, time etc., then the logarithm is “effectively bounded”, because ...

4

As others have mentioned, for all intents and purposes, yes it reaches %99 charge after 5 tau. However, as the current gets smaller and smaller as we reach full charge, technically it will never become 'fully' charged, even in practice. The current will continue to get smaller and smaller, until it is unmeasurable and therefore negligible.

3

To fully charge a capacitor to 5 Volts, say, you could connect it to a 10 Volts source until it is half charged, then connect it to your 5 V source. This is of courcse a ridiculous method, since you could hardly hit the moment of correct charge so precisely; any micorvolt error would start an exponential curve as in your original setup. That being said, ...

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 ...

0

The capacitor is connected in parallel with the bleeder resistor. The voltage between the terminals of the capacitor is equal to the potential difference between the bleeder resistor. So the volatage should......?

-1

Hint: The 10 micro farad capacitor can be removed ignored since it gets stuck in a wheatstone like network.

1

You will have to use Kirchoff's law to get the answer. How can some of positive numbers be zero? No, the sum of all charges will be zero while that of positive plate will be finite. Now the net sum would be zero since charge is conserved on the system having all the right plates. Use Kirchoff's Loop Law and Kirchoff's current law to find charges and ...

1

Hard to know why you got a different answer in your calculation, but here are some general considerations. Assume you have a circuit like this: You would think that you can easily compute the time constant $\tau = R_1 C_1 = 10^{-4}s$. However, it is most likely that your real circuit isn't actually like this. First - capacitors are notoriously ...

1

There is no need for taking the mod of the charge. The voltage has the same sign as the charge. So if you start out with +20 µC on one capacitor (they give the + sign for a reason - so that's the side where we will put the positive charge) and +60 µC on the other (from $Q=CV$) then it follows that the total of redistributed charge is 80 µC as you correctly ...

0

I think this is really about which way you count current and voltage to be positive. For every element in a network you can define a a current and a voltage. If voltage and and current point the same direction, it's called "receptor". If they are in opposite direction its' called "generator". The most common convention is to use "receptor" for resistors, ...

1

The bleeder resistor is across the capacitor so they have identical voltage across.

2

I recently answered a similar question here. The ideal capacitor equation $$i_C = C\frac{dv_C}{dt}$$ assumes the passive sign convention which means that the reference direction for $i_C$ is into the positive labelled terminal. When you write $$iR = v_C$$ it is necessarily the case that $$i_C = - i$$ To see this, assume that both positive labelled ...

2

This is a common question. The issue is that the "Q" in $i = dQ/dt$ is not the same as the $Q$ that represents the charge on the capacitor. The variable $Q$ in use here is simply the charge on the capacitor. No problem. When the capacitor discharges the quantity of charge that is introduced into the circuit after a time $\delta t$ has elapsed is ...

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