Hot answers tagged electric-circuits
10
First, Field strength.
This calculation is strictly an electric potential calculation; radiation and induction are safely ignored at 50Hz.
For a 200kV transmission line 20m above ground, the max electric field at ground level is about 1.2 kV/m. This number is reduced from the naive 200kV/20m=10 kV/m calculation by two effects:
1) The ~1/r variation in ...
10
AC or DC, you only get electrocuted if current passes through your body. (Current passing through any part of your body can be dangerous, and possibly cause an electrical burn, but current passing across your heart is the one that's really dangerous.) Touching just one wire at a time gives the current nowhere much to go. You are right to think that some ...
9
If the power line is 20m high, and has the voltage of 1MV , then the electric field (near ground), very roughly, is on order of 1000/30 kv ~ 30 000 v/m (the numbers are very approximate and the field is complicated because it is a wire near a plate scenario, and wire diameter is unknown but not too small else the air would break down, i.e. spark over, near ...
7
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.
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). ...
5
I've just sacrificed an AA manganese alkaline battery to the cause of physics.
When I first shorted the battery it produced a current of about 9.5 amps, which I thought was actually pretty impressive. However over the course of 30 seconds the current dropped to around 5 amps. The battery got pretty warm, though I don't think it would have set fire to ...
5
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 ...
5
Power consumption is about linear with frequency.
The processor contains millions of complementary FETs as shown. When the input goes low the small capacitance gets charged and it will hold a small amount of energy. A same amount is lost during the charging. When the input goes high again the charge will be drained to ground and be lost. So with each ...
5
Why is current not 0 in a regular resistor - battery circuit immediately after you closed a circuit?
In real life, the current can't jump instantaneously because there is always some finite inductance in a circuit. However, this is just a typical idealized textbook problem where the inductance is assumed identically zero, so the current can jump instantaneously according to the assumptions of the problem. Note the current also jumps in their solution for ...
5
Ohm's law $\vec\jmath=\sigma\vec{E}$ can be derived rigorously in the limit of small electric fields using linear response theory. This leads to Kubo's formula for the electric conductivity, which relates $\sigma$ to the zero frequency limit of the retarded current-current correlation function.
$$
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5
When there is no resistance, as is the case with an ideal wire, any value of current satisfies Ohm's Law:
$V = I R$
since both $V=0$ and $R=0$.
UPDATE:
But isn't V is like what causes the current?
Perhaps a mechanical analogy of the resistor will help. Consider the dashpot where the velocity of the arm is analogous to current while the force acting ...
5
A battery is basically just a chemical reaction. At the negative (cathode) end of the battery the reaction releases electrons while at the positive (anode) end of the battery the reaction consumes electrons. As long as the external circuit allows electrons to flow from the cathode to the anode the reaction goes and the battery generates power.
If you break ...
4
To add to the "linear with frequency" point, there is also an additional factor. As that "dynamic power" increases, the temperature of the die will increase and this will also increase the leakage current through the millions of transistors, which will cause more dissipation (termed "static power")
There's a long Anandtech thread taking lots of values and ...
4
What makes it a good idea to use RMS rather than peak values
The rms value, not the peak value, is the equivalent DC value that gives the same average power.
Recall that power is the product of voltage and current:
$p(t) = v(t) \cdot i(t)$
For a resistor, we have:
$p(t) = R[i(t)]^2$
To find the average power, we must take the time average of both ...
4
When you say "we see that if the current doubles then the potential difference is halved," you're assuming that $P$ is fixed, whereas when you say "doubling the current should increase the potential difference" you're assuming $R$ is fixed. But in fact, it isn't possible to change the current while keeping both of these things constant.
Let us assume that ...
4
Sine and cosine waves are, physically, the most common. They are definitely the best description to what comes out of a wall socket, not because we like them mathematically, but because it's what comes out. (Note sine and cosine waves are equivalent, and choice between them is merely convention)
Even better, if we have some more complicated waveform - from ...
4
Firstly, a cosinusoid is just a sine wave which has been shifted by a phase of $\pi / 2$. So the two are equivalent ways of describing a wave.
One can describe a wave as a triangular wave, or a square wave or any other wave form. In fact, square waves are used as the input to all sorts of logical circuits and digital electronics.
But a ...
4
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 ...
4
Regarding what you consider to be a contradiction:
How can the total current be the same if a resistor is reducing current at some point in the circuit?
The current at any point in the circuit is the same because the current distribution in the circuit has reached a steady state (i.e., charge buildup is forbidden). Your intuition is telling you that ...
4
A human body may reflect and absorb radio frequencies, though not very efficiently. It may as well act as a resonance chamber for certain frequencies. For a signal of 100 MHz, the involved wavelength is 3 m, and so it is possible that parts of your body are acting slightly as a resonant chamber. (for an optimal resonance, you should have 1.5 m diameter, too ...
3
For the most part cosmic rays do nothing to consumer electronics.
This is not to say that they can't flip bits or even damage elements, but the rate for such effects is very, very low.
Radiation effects are routinely observed in electronics placed in accelerator experimental halls (where the radiation levels are at lethal-dose-in-minutes levels when the ...
3
okay, This was really cool and I got some help from my physics professors on this one (apparently I won't learn this until next semester) and to find the magnitude of the square of a complex number you take it times it's complex conjugate. So in this case $$\frac{V_0}{i \omega L +R}$$ is multiplied with $$\frac{V_0}{-i \omega L + R}$$ leaving you with simply ...
3
The primary and secondary powers are equal but not fixed.
In an ideal transformer with turns ratio $n$ (= # secondary turns / # primary turns) fed by an ideal ac voltage source $V_p$, the secondary current $I_s$ is determined by the transformed primary voltage (that is, the secondary voltage $V_s=n V_p$) and load resistance $R$ ($I_s=V_s/R = n V_p/R$), ...
3
If the load changes on the secondary, the primary voltage and/or current will change too.
Don't forget that the source connected to the primary "sees" the load connected to the secondary transformed through the square of the turns ratio.
$\dfrac{V_p}{I_p} = \dfrac{N^2_p}{N^2_s} \dfrac{V_s}{I_s}$
Assuming a voltage source driving the primary, if the load ...
3
You made the mistake of placing the ammeter in parallel. The result is meaningless, but the software isn't wrong. It is correctly simulating the circuit, but the current through the ammeter is indeterminate in a correct analysis.
Consider, all the current could flow through the ammeter, or all of it could flow through the wire next to it. There are many ...
3
There is no reason why you couldn't build a motor using superconducting magnets, or build a simpler homopolar motor using a length of superconducting wire. There would be no heat dissipated due to electrical resistance, but of course there would still be mechanical resistance due to friction on the moving parts. The energy needed to overcome this, and to ...
3
As David pointed out in his answer, the circuits are called the differentiator and the integrator respectively. They use operational amplifiers (op amps) to do this.
An op-amp isn't necessarily a "simple circuit" to build, since it consists of several transistors and hundreds of resistors and capacitors, but since it's extremely easily available as an IC, ...
3
When the idea of electricity was discovered, it wasn't known that there were atoms, and that there were electrons that could move around. Since the definition of electric charge is completely arbitrary (we just need to opposite charges, doesn't really matter which is which) they figured that it was the positive charge that flows through the circuit.
That's ...
3
No, they do not all have the same voltage drop. If they were in series, however, they would.
By Ohm's Law, the voltage drop is proportional to the current flowing through a resistor. (So in several series resistors with the same resistance, the drop across each one is the same, since the current across each one is the same). However, because B and C are in ...
3
Electrons that reach the positive terminal indeed remain there. The potential difference between the two terminals pushes electrons from the negative anode toward the positive cathode. When an electron reaches the cathode, it stays there to equalize the original charge imbalance between the two nodes. When electrochemical redox reaction sustaining the ...
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