# Tag Info

5

Imagine electricity as water in a pipe. The current can flow in either direction (direct current, DC) or one way then the other way (alternating current, AC). Now put a rubber membrane in the pipe. This is the capacitor. Now it will slow and then stop DC, but AC can still keep wobbling back and forth. In this way, capacitors block DC but enable AC. ...

3

Vinzent already explained about RC filter transfer charcteristics, which apply nicely to single frequency. I will attempt to address the "broad spectrum signal" part of your question. As you probably know, any kind of signal can be looked as a composition of single frequencies (see Fourier transform). Broad spectrum signal can (or actually any ...

0

In the equation $$X_{C}=\frac{1}{2πfC}$$ $X_C$ is the capacitive reactance of a capacitor in response to a sinusoidal voltage source of frequency $f$ in Hz. The capacitive reactance is the magnitude of the impedance of the capacitor. Basically, it is the magnitude of the capacitor's resistance to current flow. Since you have stated that you have not yet ...

2

This is very unlikely since the the water from rain comes in single droplets and not long connected streams of water. However if it did rain the way you describe and for an instant there is a stream of water going from the power lines to you or your car, then you could get electrocuted. The high voltage in power lines have the actual physical ground as the &...

6

What is the physical behaviour which allows a capacitor to act as a high or low pass filter? A capacitor alone cannot act as either. To create a filter you need a combination of resistance and capacitance or inductance and capacitance (or RL). You need two immittances, at least one of which is reactive. Let's take a practical example, an RC circuit. This ...

19

An ideal capacitor has one intuitive property: Its voltage can't change instantly since its voltage is dependent on the charge it has stored, and charge doesn't move at infinite speeds, therefore you can't instantly charge up a capacitor without infinite current. More capacitance means less voltage for the same amount of charge, like how a wide bucket or a ...

5

Sometimes for physical intuition, it's nice to think about the extreme cases. For instance, a zero frequency signal is just a DC voltage. If we send it through the RC high pass filter, the capacitor is just like a break in the circuit, and prevents any current from flowing.

0

But if we turn the transistor on, that will [sort of] allow the current to flow through the transistor here, and that will lower the voltage potential at this node. This is a correct statement in that the node voltage in question does decrease when the transistor is on. Consider: put the black lead of your voltmeter at the emitter of the transistor (the ...

1

There is a relationship between the open circuit voltage and short circuit current. It’s called the Thevenin resistance $R_{Thev}$ where $$R_{Thev}=\frac{V_{OC}}{I_{SC}}$$ Where $V_{OC}$ is the open circuit voltage and $I_{SC}$ is short circuit current. But a higher open circuit voltage for the same source resistance will give you an increase in short ...

1

The Joule integral is actually used to characterize fuses. There are two extremes for specifying the current-carrying capability of a conductor. a) Over a long time, all heat is lost to the environment. The current is given for some permitted temperature rise. Typically this is the temperature rating of the insulation. b) Over a very short time, no heat is ...

1

Rewrite that integral (which defines a mathematical operation called convolution) by performing the substitution $\tau\mapsto t-\Delta t$, $\mathrm{d}\tau\mapsto -\mathrm{d}(\Delta t)$: $$V(t) = \frac{1}{RC}\int_0^\infty V_\mathrm{i}(t-\Delta t)\mathrm{e}^{-\Delta t/RC}\mathrm{d}(\Delta t).$$ You can now see that $V_\mathrm{i}(t-\Delta t)$, with $\Delta t$ ...

3

I can see why you would find this statement opaque. Here is my take on it. When we perform an integral, we are summing up a time series of events for infinitesimal slices or steps of time. this means that the integral contains (across its start and stop limits) an accounting of what happened right at the start of the summation, the middle, and the final ...

2

In 1., how does the reading stay the same if some resistance is removed in the branch opposite to the ammeter $A1$? The reading stays the same because the voltage across the parallel combination of $R3$ and $R4$ in series with the ammeter is the constant voltage of the ideal battery. And in 2., why exactly will the reading decrease? The parallel ...

1

Your graph has a period of 30 milliseconds. Use that to find the frequency which can be expressed in terms of L and C.

1

The accumulation is due to difference in electric field at two points .HINT try relating flux and electric field for charge accumulation ${E} = \frac{J}{\sigma}$ where $j$ is the current density and ${\sigma}$ is conductivity

0

Laboratory instruments have the impedance which is best for the job. For measuring voltages, this is typically high. For generating signals, this is typically low. For other tasks, other impedance are ideal. For example, in many RF signalling situations, you want an impedance that matches that of the circuit under test.

1

This is generally true. the higher the sensor impedance, the less it perturbs the circuit being measured. The lower the signal source impedance, the better able it is to drive a variety of circuits. Ideal sensor impedance is 10 million ohms; typical source impedance is from 50 to 600 ohms.

1

When electrons zip around in a wire, they occasionally collide with the atoms of the wire. As a result, the electron can impart some (or all) of its kinetic energy to the atoms of the wire. This causes them to vibrate. And it’s this vibration that is macroscopically seen as the heating up. Consider the atoms in the dielectric. They can be thought of as ...

1

Current trough a wire gets the electrons moving, ( I would not talk about traveling energy) they are slowed by bumping in atoms and get them to move or vibrate faster, which is heat. The stuff between the plates of the capacitor ist partly polarized, so the electric field has shorter ways, the same thing as getting the plates closer zigether.

1

The constitutive relationship of an ideal capacitor is $$i = C\frac{\mathrm{d} v}{\mathrm{d} t}$$ when voltage and current directions associated to the capacitor are chosen according to the passive sign convention, regardless of whether the capacitor is charging or discharging. If, instead, you chose the active sign convention, the constitutive relationship ...

1

DC electric motors do not produce smooth torque as they rotate; the torque has a rising and falling component that depends on the angle between the field lines produced by the field magnets and the plane containing the wire loop in the armature which happens to be energized by the commutator/brush assembly at any given moment. This means that DC motors ...

1

That 1k Ohm confuses me. Is it serial or parallel to others? Ask yourself, how much current flows through this resistor? With that in mind, how does it affect the rest of the circuit? Can you tell me about how to determine the equivalent resistance of circuits with capacitors in general? In general, you apply a test current into or test voltage across the ...

4

How then can we define a potential difference across a resistor with time varying current? Basically we just assume that we can. Circuit theory is an approximation to Maxwell’s equations which relies on three assumptions: the distances are small enough and the time scales large enough that we can treat electromagnetic effects as instantaneous rather than ...

2

When we say "current will flow" and "current won't flow" we are using phrasings which let us simplify the circuit. By using logic we can deduce that current will not flow through the 4 ohm lamps. However, if we are not so sure, we can just calculate. We can assume some current $i$ goes through the 4 ohm lamps, and solve the equations. ...

1

No, adding a resistor there won't cause the 4 ohm lamps to light up. You need a potential difference across the lamps. One way to achieve that is to move your resistor so that it's between the two 4 ohm lamps. In these introductory problems, we normally assume that the wires are ideal, with zero resistance. (And we ignore the internal resistance of the ...

1

TL;DR: The battery has to supply twice the energy that gets stored in the capacitors. The characteristics of your power source matter. The "three capacitors in series" don't make a difference, they behave just the same as one capacitor of appropriate value C. In your question's title (but not in its body), you mention the power source to be a ...

1

The energy conservation law tells us that the energy stored in the capacitors will be the energy supplied by the buttery, minus the energy dissipated in the wires while charging the capacitors.

0

The question is a multiple choice and one of the answers state 'power dissipated in the battery is 18W'. This (C) is the correct answer as the total resistence in series is 4 ohms, the current through the load is 3A not 6A and the PD across the load is 6V.

0

If the loop is going down at a velocity $v_y$, for a magnetic field B, loop width $L$, and portion of loop height in the field $= y$, $$\frac{\partial \phi}{\partial t} = \frac{B \partial (Ly)}{\partial t} = \frac{LB \partial y}{\partial t} = LBv_y$$ As you said, an electric field is generated in the conductor:$$E = -\frac{\partial \phi}{\partial t}$$ If ...

0

Firstly, you show some confusion between $Q$ an $q$. They are effectively the same thing, the charge stored by the capacitor. Since it is changing, it is customary to use lower case. $dq/dt$ is the rate of change of $q$. As the capacitor discharges, $q$ falls; thus its rate of change $dq/dt$ is negative. However the current $i$ flowing out is a positive ...

0

The current from $V_2$ is already accounted for by $I_3$. When you solve the circuit you will find the current coming from the + terminal of $V_2$ is equal to $-I_3$. Also, it is possible for a voltage source to be absorbing energy (current flow positive into it). You can assume current directions when you draw your arrows - if you assumed wrong direction ...

4

Here is something to get you started. Firstly let's make a simple SPICE model for a single solar cell. We have a current source in parallel with a diode, we also have a voltage source which we will sweep to calculate the IV curve. Here is the spice model, * Single solar cell solar cell Isc1 0 1 390.30 d1 1 0 GaAs_diode vs 1 2 0 vcc 2 0 0.7 * Ideal GaAs ...

6

Well if there is a wire coming from the positive terminal of the battery, then into the remote’s circuitry then to the negative terminal, water may cause a link between these wires before the current has a chance to circulate through this circuitry. This will obviously stop the remote from working at all. Sometimes the water will short other component wires “...

1

In sinusoidal steady-state (linear loads, no harmonics), the meaning of reactive power is known. In non-sinusoidal steady-state (non-linear loads, harmonics), I haven't read a textbook or article, or seen a video, or heard someone explain it; even literature doesn't have a unique definition of reactive power in these conditions, let alone its physical ...

0

Potential energy is always defined relative to some reference position. Unless your circuit is “grounded”, you can choose that point (connect the black lead from your voltmeter there). Then, the potential at any other point in the circuit is defined (and measured) from there.

0

I have been told that potential in an electric circuit is defined as the work done in bringing a unit test positive charge to that point inside the circuit. True, but you have to know where the point is in the circuit where the test positive charge starts. Then, the potential difference (voltage) between the two points in the electric circuit is the work ...

0

In this situation, there will be a magnetic force acting on the free electrons in the upper segment of the loop, pushing them to the left and producing a potential difference, vBl, between the two ends of that segment. On the other hand, a closed conducting loop totally within a uniform magnetic field which is changing with time will experience an emf which ...

1

No proof exists for the combined relation: $I=-\frac{d \Phi}{d t}/R$ ...because it is not true in all cases. This happens because the current induced in the loop generates it own internal magnetic flux that opposes the external magnetic flux according to the Lenz Law. That reaction modifies the variable $\Phi$ in the equation above. Consider the limiting ...

3

Good question. In my experience, most introductions to E&M don't give a great explanation of the exact similarities and differences between voltage and electromotive force (emf), and when you can and can't use the concepts interchangeably. Voltage and emf are both formally defined the exact same way, as the (negative) line integal of the electric field ...

1

You say"potential in an electric circuit is defined as the work done in bringing a unit test positive charge to that point " You do not say from where you bring the charge, i.e. where you have potential zero. Usually one attributes potential zero to one point of the batterie, or to some point connected to earth.But this has nothing to do with the ...

1

If you draw an RC circuit without generator, and you use Kirchhoff laws, you get that the tension across the capacitor goes to zero with an exponential function with a time constant $\tau =RC$. This means that after $5 \tau$ the tension is zero for practical applications. You said long time, this is relative. For example if you have R=1KOhm and C=10^-6 F you ...

1

Go through Phasors. Phasors are used to represent a time-domain signal $f(t)$ as a frequency domain signal $\hat{f}(\omega)$ so as to solve the integro-differential equations quickly. We have, $\exp[\pm i(\omega t+ \phi)] = \cos(\omega t + \phi) \pm i\sin(\omega t + \phi).$ So, $\cos(\omega t + \phi) = \mathfrak{Re}\{\exp[\pm i(\omega t+ \phi)]\}$ and $\... 1 The potential difference,$V$, between the plates of the capacitor is also the potential difference across the resistor, because the resistor is connected across the capacitor (assuming that this RC circuit consists just of the R and the C). The pd$V$will drive a current ($I=V/R$) through the resistor. This current can only be a flow of electrons from the ... 0 It could very well be the current was pointed in the wrong direction. It does not matter, however, because after calculations you will see if your guess for the direct current was correct. If you get a negative sign for your current value, it means that you chose the opposite orientation. Also, the way the loops are oriented is due to the direction of the ... 0 If your circuit has two “holes” through it, you need to define only two current loops (which may add (or subtract) in some parts of the circuit. Then you can write and solve two voltage loop equations. Your choice of current and voltage loops is arbitrary as as long as all parts of the circuit are included. 1 how do I calculate how much voltage and/or time I need to put a given charge on the material? If the material is not conductive, then it is difficult to put the charge across the material. Since charging up one tiny point doesn't really affect the rest of the material. So "how you do it" would be critically important to understand how long it ... 1 You charge an insulator (e.g. methyl methacrylate a.k.a. Plexiglas) by targeting it with a cathode ray. The abstract linked below quotes a value of 2 million volts. I remember this sort of thing was done in the ion implantation lab where I worked as a student years ago. Those voltages were lower, probably no more than half a million volts. https://www.... 0 OK, you mean calculate. (You would measure it with a spring balance - easy). You want to start with$H$rather than$B$, but that's just a factor of$\mu_0$. Then you need the magnetisation$M$. For dia/paramagnets this is just$\chi H$, where$\chi\$ is the magnetic susceptibility. For ferromagnets it's more complicated due to saturation and hysteresis. If ...

-1

A current carrying solenoid will exert a force on a ferromagnetic material which is just outside the end of the solenoid. A calculation would be complex. The magnitude and direction of the force on each small segment of the material will vary from point to point and depend on the divergence of the field and the existing or induced magnetization within the ...

0

Use the expression for the Lorentz force. If the object is a punctual object (i.e. a charged particle) the Lorentz force is $$\vec{F}=q(\vec{E}+\vec{v}\times\vec{B}).$$ If the object is extended then you'll have to integrate $$\vec{F}=\int(\rho\vec{E}+\vec{j}\times\vec{B})d^3x.$$ Notice that the object needs to be charged in order to experience an ...

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