# Tag Info

1

I don't understand why the resistor would have have all the voltage and why wouldn't the capacitor have the voltage. Is this because the capacitor is being charged 'through the resistor'? And what does this mean? The short answer is the voltage across an ideal capacitor cannot change instantaneously, i.e., in zero time. It therefore looks like a short-...

0

The potential drop across a capacitor is directly proportional to the excess charge stored on each plate of the capacitor: $V_C=q/C$. If there capacitor starts off discharged then it has no potential drop across it. By Kirchoff's loop rule, this means that initially the potential drop across the resistor has to be equal in magnitude to the potential across ...

1

If switch is opened when capacitor is fully charged energy of LC system remains same. If it's a series LC circuit, then the energy will be $E=\frac{CV^2}{2}$ all residing in the electric field of the capacitor. If it is a parallel LC circuit of an ideal inductor and capacitor, the energy will remain the same but it will be exchanged between the electric ...

2

When the capacitor has no charge, the energy is in the magnetic field of the inductor, which is associated with a current flow. If the switch is open, the current cannot flow. The magnetic field collapses, leaving no energy. (The collapse of the field will cause a large voltage spike, and probably an arc across the opening switch. The energy is dissipated ...

1

I'm not sure how you're supposed to calculate the time constant of a capacitor when a direct current is applied. The time constant for the capacitor is simply RC and it applies to both AC and DC circuits, but only under transient conditions such as during a period of time just after a switch connects or disconnects a capacitor to the circuit. After a long ...

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According to one of my professors: I am not 100% sure, but I guess for the ideal diode it should be something like $$4 E_C \hat{n}^2 H(\hat{n})$$ where $H(\hat{n})$ is the Heaviside step function. If the charge is negative, the capacitor is immediately discharged and the energy of the system is $0$. If the charge is positive, the energy is just the energy ...

1

I have never seen a Hamiltonian for a circuit with a diode, and I doubt that it exists or used - for the reasons I describe below. However, from purely academic perspective it is an interesting question to ponder. Unlike inductance and capacitance, which can be characterized by linear response and therefore described by quadratic Hamiltonians, a diode is a ...

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I don't think you would need to integrate radially inward specifically, but what is required is to integrate radially. You can take $dr=ds$, this will mean you are integrating along the field, i.e. outward instead of the opposite that is done in the book. Here's the thing if you integrate by taking $dr=ds$, the final result is $$V= \frac{-q}{2 \pi \epsilon ... 1 This is a nice question, Consider this circuit first, that build from one resistor and battery, We can apply kirchhoff's law and we can get that$$V_R=V$$while V is the battery voltage, thus we can say that the resistor is connected parallel to the battery. Now check the following circuit, We can see that applying kirchoff's law here will yield$$V= V_{...

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I assume that this is an example where one charged capacitor charges another after the switch ic closed. The main use of assigning the labels series or parallel to capacitors (and other circuit elements) is to decide which combination rule to use to find the effective capacitance of a number of capacitors. The derivation of such combination rules have ...

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They are in series, one end of the first capacitor contacts one end of the other. They are also parallel when the switch is on, because they both connect two ends. Parallel connection means both ends of the two elements are connected together. This happens when you turn the switch on. Both ends of C1 becomes connected to both ends of C2. Before that, only ...

2

My question is given that a capacitor creates a two charged sides, by the electrons jumping from one plate to another thereby making one having an excess of electrons making it negative and the other positive. If you mean by "electrons jumping from one plate to another" that the electrons move across the space between the plates, that is not the ...

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You have got the right idea. When a circuit consisting of a battery and capacitor is switched on, the electrons from the negative terminal start accumulating on the capacitor plate. The negative capacitor plate induces an equal and opposite charge on the other capacitor plate. Together, an electric field is created and there is a new potential difference in ...

-1

Energy stored in an electric field - Means the Potential Energy (electric) in that space. You do not even need to know volume for energy stored in electric field. It has three equations. PE = (1/2) C[V(net)^2] where C is capacity and V is 'electric potential'. I am sure you can find the other two online. C has units of F or Farad V has units of Volt or ...

0

The maximum charge will depend on the material separating the plates and other parameters. If there is nothing between them but air or a vacuum then that will determine the maximum charge. The capacitance of parallel plates is given by: $$C = \frac{k \epsilon_0 A}{d}$$ where k is the relative permittivity of the material between the plates. So the maximum ...

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If the accuracy of scope of somebody is better than the accuracy of my scope. For example, if each jump of the cursor of scope is 1µ s ( Δ x/ Δ t), the angle would be 89.99 ° or 90 ° . (Each jump of my scope is 4µ s) Besides, my method to find θ and Φ is (1) using high pass RC circuit to find θ (2) then, using low pass RC circuit to find Φ (exchange the ...

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I used the phasor diagram of RC circuit to analyse the experiment. If the angle between Vr and Vs is θ and the angle between Vs and Vc is Φ . The addition of θ and Φ should be near to 90 ° ( Vr : voltage of resistor; Vs : voltage of source; Vc : voltage of capacitor; I : current ) Then, I found the angles of θ and Φ in the experiment. Since Vr and I are in ...

0

I have designed a simple experiment to show that it is true within the experimental errors and wanted to know whether there is another method to show that it is true. (One of the experimental errors arises from the accuracy of my scope. ) I used a series RC circuit. R is 22k ohm. C is 0.01 micro farad. Source is 4Vpp with 500Hz/1000Hz/1500Hz Sine Wave. My ...

2

Use an oscilloscope. Let one channel display the voltage across the capacitor. Let a second channel display the voltage across a resistor, which you have wired in series with the capacitor.

0

You connect the battery which puts a net charge $+Q$ on the top plate and $-Q$ on the bottom plate. The plates are attracted to each other and when you pull them apart you are doing work on the system. My guess is that this work ends up forcing some of the charge on the plates to flow back to the battery. Maybe try this calculation: Separate the plates by ...

0

The electric field on one plate is "felt" by the other. A simple way to think about why the distance between the plates matters, is that the closer the plates are, the more strongly will the field of one plate help pull charges towards the other plate. Thus, more charge can be accumulated on the plates when they are closer. When the charges are not ...

1

When the separation distance $d$ increases, all other things being equal, the capacitance is less, as you already know. But the charge doesn't "go anywhere". You don't "lose charge". With no voltage applied to the capacitor, the charge on the metal plates consists of an equal number of mobile free electrons and protons. The total charge ...

3

$\frac{9dQ^{2}}{10\varepsilon_{0}A}$, and I claim that all of this energy would be wasted on the resistor. This is where you go wrong. Some electrostatic potential energy will remain in the circuit. The only way there would be no remaining potential energy is if the voltages on both capacitors was zero after a long time. But that cannot happen due to the ...

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