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You can use a high vertical tube to store water in it (fill it from the bottom by pushing the water in) How much water can you store? It obviously depends on the pressure you apply to push it in. If you push harder, there will be more water stored. The tube is characterized not the amount of water, but by how easy it is to store the water. Its "capacity" ...


4

Capacitance is "charge over voltage" – and one farad is "coulomb per volt" – because the capacity of capacitors (something that determines their "quality") is the ability to store a maximum charge on the plate ($+Q$ on one side, $-Q$ on the other side) given a fixed voltage. When you try to separate the charges, you unavoidably create electric fields ...


4

a. Immediately after the switch is closed, are either or both bulbs glowing? Explain. They will both glow as some current passes through them as the capacitor is charging. b. If both bulbs are glowing, which is brighter? Or are they equally bright? Explain. They are both equally bright, because an equal and opposite charge is flowing on to ...


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First answer this: does it take more power to run one light bulb or two? Transmission lines are designed to be low loss, but they run a long way. Lightbulbs are designed to be "lossy" because that's how they work. Let's say for the sake of argument that the whole transmission circuit loses 100 watts and you're talking about a 100 watt bulb: those numbers ...


3

While the other two answers are technically correct, they are not actually addressing the engineering aspects of what comprises a "good" capacitor. Ideal capacitors in parallel or series circuits lead to ideal capacitors of different value, i.e. there is no measurable "quality" difference - ideal is ideal. In reality, of course, there are no ideal ...


2

To this EE's eyes, both formulas are incorrect; the reactance of a capacitor is $$X_C = -\frac{1}{\omega C} $$ and so the sign is incorrect in the first formula. Since reactance is the imaginary part of impedance $$Z = R + iX$$ reactance is a real number and so the second formula can't be correct. It is instead the formula for the impedance of a ...


2

If you have a known load $R'$, you can divide the potential of a source with a single resistor. If your source voltage is $V$ and you want a voltage $V'$ across the load, you can write $$V' = \frac{R'}{R_L+R'}V$$ from which you solve for $R_L$, the load. However, if you don't know $R'$ (or if it can change - for example, as a light bulb heats up, its ...


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The only lossy component in the circuit is the resistor and the power dissipated in the resistor is $I_{\text{rms}} R$ so the conversion $I_{\text{rms}} = \dfrac {I_{\text{peak}} }{\sqrt{2}}$ has to be made assuming that the variations of current and voltage are sinusoidal. The mean power dissipated in the capacitor over a cycle is zero.


2

if we connect the set up like this the bulb glows If the pin positions in the photo are Ground Neutral Line And you are in a country where the neutral is linked to ground at the main panel but where protective devices such as GFCI or RCD are not used. Then you have a path for current to flow from Line via Ground to Neutral. if ...


2

(1) If the wave is induced by and propagation from the voltage source (battery), then it should take the vector path of the magnetic field created by the battery, instead of the circuit path. Yes, and the magnetic field follows the circuit path, but remains outside the wires. The electric current in a wire is causing a magnetic field which is located ...


2

For example in simple circuit with lamp. I understand that the energy is spent for heating and lighting. But how exactly it happens? The MIT Teal project has an excellent short video which can be used as an animated diagram of the (Classical physics) process inside a resistor such as a tungsten filament. In the video below, imagine that the vertical ...


2

Note that the sum of two phasors is another phasor: $Ae^{j(wt+\phi_1)} + Be^{j(wt+\phi_2)} = Ce^{j(wt+\phi_3)}$ Where A,B,C are real. The only way for the real part of the right side to be equal to zero at all times is if $C=0$. In which case the whole thing is $0$, both the real and complex parts are 0. So a sum of phasors (of the full sinusoidal form ...


2

The current doesnt pass through the voltmeter so the the resistance of load R is not affected.


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The idea of the linked page is this: you can often find the equivalent resistance of certain highly symmetrical assemblies of resistors by adding "phantom" wires between points with equal potential. This has the advantage of giving quick, near-automatic answers without the use of Kirchhoff's laws, but it is very limited and works in very few cases. This ...


2

It is sometimes easier to visualise what is happening by using the idea of potential. To do this make one point in the circuit 0 V. This is a totally arbitrary choice. It is the bottom right hand corner of your circuit. Note to make the sums easier I have change the emf of the battery to 90 V so 2 A flows through the battery and 1 A through each of the ...


2

When you apply an AC voltage to a capacitor, current will flow into the capacitor, and back out again. As long as the absolute voltage on the AC generator is higher than on the capacitor, current will flow to increase the charge on the capacitor; and when it's smaller, current will flow to decrease the charge. Note that since there is an equal and opposite ...


2

For copper the temperature coefficient of resistivity is $3.9\times 10^{-3} \text{K}^{-1} $ and the temperature coefficient of thermal linear expansion is $1.6\times 10^{-4} \text{K}^{-1} $. They differ by a factor of about 24 so a change in temperature will cause a bigger change in resistance than in the linear dimensions of copper. Resistance is given by ...


2

Georg Ohm's original experiments, 1825, established that for a set temperature, the current through a specific length of a conductor was proportional to the potential difference applied. Ohm's law is empirical; it cannot be derived directly from Maxwell's equations as it depends upon material properties. It is violated by many materials, and even then ...


2

A capacitor is used to store energy in form of electric fields. This electric field is created by charges on plates of capacitor. So, basically you are storing charge on capacitors. Let someone ask you how much charge you can store in your capacitor.What would you reply? Clearly , you reply " I may store 1mC or 100mC, depending on Potential difference ...


2

I believe the main effect of Earth Hour is in attracting public attention to sustainability. But besides of it, without doubt it saves energy (... and natural resources as well as one's wallet) if one turns off their unused light. The scale of one hour savings on turning off dispensable lighting for one hour is however rather small compared to other means ...


2

We Use $C=Q/V$ because those were useful things to measure. It's often easy to forget, but many of the equations we use are chosen because the work, and because other equations didn't work. Never underestimate that part of the reality. We don't use "charge per unit volume" because that number is not constant. You can charge a capacitor up without ...


2

I understand that capacitance is the ability of a body to store an electrical charge and the formula is $C = {Q \over V}$ Perhaps you just need to top thinking of capacitance as that. "Capacitance" sounds like "capacity", which leads to an intuitive trap like this: If I have a basket with a capacity of 2 apples, then a basket with more capacity can ...


1

For instance, why don't measure the ability to store something by the volume it takes so why not charge per unit volume. There is nothing wrong with you defining a parameter which is the "charge per unit volume" but after defining it then what are you going to do with it? So here you have a capacitor and its charge per unit volume is $3 \;\text{C ...


1

I want to charge a 12v 100ah battery which need 20 amp of current. You can charge the battery at any current proving it is not too high and it might be that the 20 amp is the maximum charging current? If the adapter gives a constant (regulated) 12 volts then you will not be able to charge a battery of the lead-acid type it will require more than 12 V ...


1

If the inner conducting shell is charged then the charges will reside on the outside of the inner shell as there can be no electric field inside a conducting shell. The charges on the outside of the inner conducting shell will produce a radial electric field and the outer conducting shell will find itself in that electric field. However the outer ...


1

You're not the first, nor the last, to find the phrase "power flow" somehow wrong. For example, from W J Beaty's article on electrical misconceptions: ELECTRIC POWER FLOWS FROM GENERATOR TO CONSUMER? Wrong. Electric power cannot be made to flow. Power is defined as "flow of energy." Saying that power "flows" is silly. It's as silly as saying that ...


1

Adding another battery in parallel could allow the supply of more current, but if the current you need (for your circuit and battery voltage) is already sufficiently supplied by the first battery then you won't see any difference. Note also, that in generally you shouldn't connect batteries in parallel, and it can even be damaging or unsafe to do so. For ...


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I'm not sure exactly what you're asking for. But let's say there's an electric current flowing through a straight wire segment of length $l$, then the change in $\Delta\phi$, or $V$, would be defined by $$\Delta\phi = \int \mathbf{E}\cdot d\mathbf{l}$$ Because it is a straight wire, $$\Delta\phi = E\cdot l$$ But we have a definition of current $$I = ...


1

why would increasing voltage, while keeping charge constant, have any effect on the ability of a body to store charge. (1) Capacitors don't store charge, they store electrical energy. For a capacitor, it is understood that one plate has charge $Q$ while the other plate has charge $-Q$ so there is no net electric charge stored. (2) If you increase ...


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Power consumer and power dissipated will be different. Power consumed takes into account power supplied to the resistor and the capacitor. Power dissipation is the I^2.R losses due to heat. The power consumption of the capacitor falls on the imaginary axis. Hence we never observe it as heat. Here's a guy asking a similar question: ...



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