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Consider the instantaneous current $i(t)$ through a resistance $R$. The instantaneous power dissipation is
$$P(t) = Ri^2(t).$$
The average power dissipation during a sufficiently long time $T$ after $t=0$ is
$$P_{avg}=\frac{1}{T}\int\limits_0^T{P(t)dt}=R\frac{1}{T}\int\limits_0^T{i^2(t)dt}=R\sqrt{\frac{1}{T}\int\limits_0^T{i^2(t)dt}}^2=Ri_{rms}^2.$$
This ...
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Impedance can be seen as an analogy to resistance. As a simple example, if we consider a resistor connected in series to a voltage source, we have $R=V_R/I$
We can do the same thing for a pure inductor hooked up to an alternating voltage source. If $V(t)=V_0\cos(\omega t)$ for example, then $V_L=-V_0\cos(\omega t)=L\dot I$ which then indicates that $$I(t)=-\...
3
If both bulbs have identical normal tungsten filaments which are rated to run directly from the given battery, bulb B will come on bright but quickly dim because the resistance of tungsten increases with temperature. The inductor limits the rate at which the current increases in bulb A. Its brightness will start low and come up fairly fast. Since the ...
3
Here are a few reasons:
As noted by Slereah in the comments, signals would propagate much slower: by roughly a factor of $10^5$.
Thermal conductivities of common materials vary by about a factor of $10^5$ between the most thermally conductive and insulating objects. For electrical conductivities it is $>10^{20}$. This means that you can have nearly ...
3
By definition the rms, also called the effective or heating value of AC, is the equivalent of DC with respect to resistance heating. The reason for taking the square is because both positive and negative values of current equally produce resistance heating.
Hope this helps.
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They don't technically have to touch. They just have to get close enough for the field to exceed the breakdown limit of air (around 3kV/mm).
So if you're doing something at 750V, you just need around...
$$d = \frac{V}{E_{break}} = \frac{750\text{V}}{3\text{kV/mm}} = 0.25\text{mm}$$
Once the wires are closer than this, an arc will tend to form. In practice, ...
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Is it true that an ideal capacitor will take no time to discharge?
No, it's not true.
UPDATE to address some comments (since comments are ethereal):
Is it true even when capacitor and the connecting wires have no
resistance?
Eyyboss, does the loop formed by the zero resistance wires and capacitor enclose non-zero area? If so, then there is non-zero self-...
3
Number 1 says if you add a resister in series to a circuit, the current will drop which is true. Number 2 says that once you have added the 2 ohm resister to the 1 ohm resister circuit, the current passing though each of the two resisters will be the same, which is also true. They are not saying the current would be the same as it would be with just the 1 ...
2
Actually, electrons move very slowly in a typical conductor like copper: on the order of microns per second. (See Wikipedia "drift velocity".) However, an enormous number of electrons are moving, so it takes hardly any time for enough charge to be transferred to discharge the capacitor completely. A good analogy is to think of the wires as being ...
2
Let me offer an Old fashioned answer: Voltage is NOT like push! Voltage is more like pressure in a fluid, where now “pressure gradient” causes push. Electric fluid gets pushed from high pressure to low. So voltage gradient or volts per metre gives electric field or force per coulomb. Hope this helps too.
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An ideal wire is really a limit case. Consider the limit of a resistor as the resistance goes to zero. The electric field approaches zero, but the voltage across it also approaches zero. As long as the current is limited by other components (such as other resistors), the result will approach that of an ideal wire.
If you have a short circuit, you don't ...
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I understand that Z is complex number and I understand the relation
$\tan\theta=\frac{\mathfrak{Im}(Z)}{\mathfrak{Re}(Z)}$ but why this angle is represent the phase angle
between current and the voltage?
The impedance $Z$ of the circuit is the ratio of the phasor voltage $\tilde{V}$ across the circuit to the phasor current $\tilde{I}$ through the circuit.
$$...
2
There are examples of circuits consisting of DC (constant) sources, resistors and capacitors that do no have a DC steady state solution. A simple example (with ideal circuit elements) is a series RC circuit driven by a (non-zero) constant current source. The capacitor current is then constant.
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When an electron moves towards proton the potential energy loss , is converted into gain in kinetic energy. so no heat is dissipated.
when electrons are moving in resistors due to an electric external field, it's average speed until it collides (Drift velocity) remains same to due to successive collision with lattice, so you see the kinetic energy of ...
2
The power dissipation of vacuum tubes varies greatly depending on their size and what they are designed to do. A typical vacuum rectifier (diode) will emit (as waste heat) anywhere from a watt to a hundred watts in the process of passing anywhere from tens of milliamps across a hundred volts to hundreds of milliamps across 500 volts or more.
This heat comes ...
2
In the lumped-model analysis of circuits, each line (or "wire") connecting components in the circuit has a single potential: i.e. the potential does not change along the line. This means that
$$V_{ap}=V_{qr}=V_{sb}=0.$$
So you and the third image are both correct.
Implicit in this "single potential" approximation are the following ...
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A battery (flat or not) is still a conductor and the circuit will remain complete. Its just that when the battery runs out, it wont be able to provide any extra voltage to the circuit. The bulb will still glow but the brightness will change (depending on if the batteries were in series or parallel configuration).
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In electronic devices of modern manufacture, the three hardware component classes that actually degrade with age are 1) batteries, 2) electrolytic capacitors, and 3) wire-to-board interconnects.
Degraded batteries exhibit progressively less charge capacity and eventually will not take a charge at all, but this would not in and of itself cause the device to &...
1
The third image isn't misleading. $C_1=\dfrac{Q}{V_{pq}}$ and $C_2=\dfrac{Q}{V_{rs}}$ are correct and also don't contradict $V_{ab}=V_{pq}+V_{rs}$ as follow
$$\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$$
$$\dfrac{V_{ab}}{Q}=\dfrac{V_{pq}}{Q}+\dfrac{V_{rs}}{Q}$$
$$\therefore V_{ab}=V_{pq}+V_{rs}$$
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Circuits are approximately devoid of node capacitance (i.e. unlike the
sphere atop a van de Graaff generator, they are presumed to hold negligible
net charge), so the voltmeter (in circuit-theory approximation) would read
zero. If the 'isolated' circuits have a voltmeter wired in, of course, they
aren't isolated from each other any more.
Even without ...
1
Great question. An ideal voltmeter with infinite series resistance would probably measure a non-zero potential difference. However, a real voltmeter will have a large (but finite) series resistance, so what you would actually see is a potential difference that is probably non-zero initially, but that decays to zero roughly exponentially.
First, picture the ...
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I don't think you can write the potential of a single plate as $\frac{\text{charge}}{\text{capacitance}}$ because how would you define the capacitance of a single plate?
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In microwave circuit analysis, it is standard practice to introduce voltages, currents, impedances, etc., such that Kirchhoff's laws (1&2) hold. This is based on the observation that waveguides sustain a well-defined discrete set of propagating modes. Take, for example, a junction to which one or several waveguides are connected and launch a wave in one ...
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I can't help but imagine about the free electrons in that wire, does
the electrons of negative terminal repel them and in that way we say
the electron move towards positive terminal?
The electric current is the movement of free electrons in response to the application of an electric field by the battery. The electric field applies a force to the free ...
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It kind of depends on what you mean by a DC circuit. If it involves e.g. a DC voltage source that is instantaneously connected to an LC combination of an inductor and a capacitor with zero-resistance wires, then no, a steady DC state is never reached -- but presumably that's not what you meant.
If, instead, you have a graph of capacitors and resistors (with ...
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Thanks for submitting your idea! I think "charge" means two different things in your two different equations, and so you can't substitute one for the other. Coulomb's Law comes from electrostatics, where it talks about a single electric charge, and to get the total voltage, you sum over all charges. The integral of current, on the other hand, ...
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I think your definition of the mean value of $f(x)$ over an interval $ab$, $\frac{\int^b_a|f(x)|dx}{b-a}$, is just as valid as $f(x)_{rms}=\sqrt{\frac{\int^b_a (f(x))^2dx}{b-a}}$. The values of both definitions though are different and have different physical units. For an AC-circuit only the second definition gives the right value and unit for power ...
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it is the ac voltage source which is creating the trouble since the direction of voltage is reverse according to its time period so if we have a dc source then power dissipated through r will be $$P=i^2r$$, so here we have to replace this I with the average current because the net current is zero in the total time.
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if $\omega$ is the frequency of an ac signal, then the inductive and capacitive reactances are given by
$X_C = \frac{1}{C\omega}$
$X_L = L\omega$
You know that at resonant frequency, $\omega_0$, both $X_L$ and $X_C$ are equal. Now if we increase $\omega$,
$X_C$ decreases and $X_L$ Increases from the same value. Opposite is true when we decrease $\omega$ from ...
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but I just do not understand what does it mean? because there is no
current between the capacitor plates according to my understanding
It's true that (ideally) no electric charge flows between the plates of a charging or discharging capacitor.
And, as the accepted answer by BioPhysicist points out (at this time), it's also true that the flow of charge onto ...
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