# Tag Info

16

It depends on the internal resistance of the source. Fist consider a "voltage supply". What does "voltage supply" even mean? A voltage supply is supposed to output a fixed voltage no matter what we connect it to. Is this even possible? Suppose we connect the two terminals of the voltage supply together through a piece of wire, i.e. a really low resistance ...

11

If you express power loss in a power line as $V^2/R$, the $V$ in that expression is the voltage difference between the two ends of the power line, not the voltage difference between the power line and ground. To supply a fixed amount of power $P_L$ to a load, if the voltage at the load $V_L$ is larger, the current $I=P_L/V_L$ can be smaller. If the power ...

5

The instantaneous power expended in a resistor is proportional to $V^2$ (i.e. independent of its direction), so the average power expended is given by the mean square voltage! The square of the average absolute voltage will not yield the power expended in the resistor. Edit: And actually this close-to-duplicate question does have more extensive answers ...

4

know that the two conversion coefficients are close but I simply cannot see why the RMS value is the one that conforms to reality. It's true that we can ask for the average of the magnitude of a sinusoid over time and calculate it. This is a useful quantity if, for example, we want the time average voltage at the output of a full wave rectifier ...

4

The answer is simple, the resistor doesn't know what voltage drop to provide, and to some extent it doesn't "care" either (unless it's so high that the current fries it but that's another issue). The "voltage drop" is only governed by the potential difference created by the generator (battery or other). If the circuit is open, the electrons have no path so ...

3

Think about current flow. If we take each individual resistor and determine the current for the applied voltage, we get: $$I_T=\frac {V}{R_1} +\frac {V}{R_2} + ...$$ Dividing everything by the voltage give us: $$\frac {I_T}{V}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Which is the same as: $$\frac {1}{R_{eq}}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Since there ...

3

The individual resistances are all positive, so the sum $$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots \,,$$ is larger than the inverse of any of the individual resistances, and that means that the inverse of the sum is necessarily smaller than any of the resistances. No mucking around with the two-resistor form required.

3

It's not a dumb question. The electrons at the negative side of the battery have energy, and will look for "any path" to get rid of that energy - move in the direction of an electric field. When you create a circuit, you move the potential of the positive terminal closer to the negative terminal. If you assume that "wire" has no resistance (for simplicity) ...

3

It depends on if your power supply is constant voltage or constant current. Usually, it's the former, so it means that the $P = V^2/R$ is the more appropriate one to use. Therefore, a smaller resistor will dissipate more power in this situation. In some situations (electromagnets is one that comes to mind), the load is driven with constant current, so ...

3

Let us crudely imagine the voltage source as a pump pumping water up to the top of a water slide, and the resistor as the slide itself with water flowing down through it. The height difference between the top of the slide and the bottom of the slide is the same as the height difference between the top of the pump and the bottom of the pump. The voltage ...

3

The definition of current density is $J = \frac{I}{A}$, or more precisely, $J = \frac{\mathrm{d}^2 I}{\mathrm{d}^2 A}$. It is always true, by definition. $J=\sigma E$ is a different equation: it's equivalent to Ohm's law, which you know better as $V = IR$. Ohm's law is not universal; it only works for certain materials, called ohmic materials. For an ohmic ...

2

The electric field strength can be represented by the equation $$E=F/q$$ where $E$ is the electric field strength in N/C, $F$ is the force in Newtons, and $q$ is the charge in Coulombs. If we take the equation that you provided: $$\Delta p=qEt$$, and substitute $F/q$ in for $E$, then $$\Delta p=q\frac Fq t$$ which simplifies to $$\Delta p=Ft$$ This looks ...

2

Waffle's answer shows you exactly why the RMS isn't the average: Here, the average value of $V(t)$ clearly isn't $2V_p/\pi$ that you've obtained, it's zero. The half-cycle business doesn't make much sense given that if you chose $\pi/2$ to $\pi$, instead of 0 to $\pi/2$, you'd have a negative average, so which would be the true "average" $+2V_p/\pi$ or ...

2

There is a simple picture that might help you: think of a wire as a pipe full of electrons. If you put pressure on one end of a pipe, the liquid in the pipe will transmit that pressure all the way to the other end. If you pump a little bit of liquid into a pipe, an equivalent amount of liquid will come out at the other end. It won't be the same liquid, ...

2

The definition of RMS is Root Mean Square and means the square root of the avarage of the square of some quantity, that is $a_{RMS}=\sqrt{\langle a^2 \rangle}$. So for the case of a sinusoidal current, we would get $$V_{RMS}^2=\frac{1}{T}\int_0^T\left(V_p\sin\frac{2\pi t}{T}\right)^2dt=\frac{V_p^2}{2}\implies\\ V_{RMS}=\frac{V_p}{\sqrt2}$$

2

Short answer: even in non-uniform fields, the speed won't change, but the guiding center can drift with some velocity. In a magnetic field (uniform or otherwise), the force on a charged particle is: $$F = q\; \vec{v} \times \vec{B(x)}$$ The direction of this force will be tangential to the velocity vector because of the cross product (the result of the ...

2

We can prove it by induction. Let $$\frac{1}{R^{(n)}_{eq}} = \frac{1}{R_1} + \cdots+ \frac{1}{R_n}$$ Now, when $n=2$, we find $$\frac{1}{R^{(2)}_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \implies R_{eq}^{(2)} = \frac{R_1 R_2}{R_1+R_2} = \frac{R_1}{1+\frac{R_1}{R_2}} = \frac{R_2}{1+\frac{R_2}{R_1}}$$ Since $\frac{R_1}{R_2} > 1$, we see that $R^{(2)}_{eq} ... 2 A general answer which is not of any particular use is that electrical energy, and the forms in which we store it, are typically very low entropy systems. The lower the entropy the more they "want" to dissipate and the harder it is to stop that tendency to turn into (ultimately) heat. Same way that it is a lot easier to store water that is 10 degC above ... 2 The internal resistance of batteries is caused by a number of different mechanisms. A major contribution comes from the ionic conduction mechanisms in the electrolyte solution. Ions are large and can only move very slowly in electrolytes. Another source for the voltage drop is the concentration and current density dependent polarization of electrodes. To ... 2 The safety of a low voltage DC power supply is not established by the voltage on its output, but by the isolation between its input and output terminals. For example, a defective 12V power supply may have a short between the 120V AC input terminal and its negative output terminal. A user who would be connected to ground would then experience a 120V AC ... 2 This is as much a chemistry question as it is an electricity question, because batteries are chemical devices. A battery is constructed such that there's a redox reaction split into two half-reactions which must move electrons through your circuit in order to complete. The reaction happens at a finite speed, and may also be limited by how fast various ions ... 2 Yes indeed. Assuming that by storing static electricity you mean storing an electric charge directly then there is currently a lot of interest in devices called supercapacitors that do exactly this. In particular they are being investigated for use in electric vehicles. Storing a high charge density in a capacitor is hard because it produces a very strong ... 2 An intuitive tool to understand resistivity in the classical context, is Drude model. Consider a 3 Dimensional lattice, of stationary obstacles, that are separated by a characteristic length$\lambda$. Now consider a charged particle, operating under a potential difference that creates a constant E-field in this 3D lattice, let's say in the$\hat{x}\$ ...

2

CuriousOne's comments basically answered your question. I will add that if enough current is allowed to suddenly flow it can vaporize materials very rapidly, including metal. This sudden vaporization can create a rapid expansion and if that expansion is restricted by something then it can explode in the same fashion as a bomb. You often hear of ...

2

how can this be true? If we connect 1.5V cell to a 10 ohm resistor, the current is, by Ohm's law, 0.15A and the power delivered to the resistor by the cell is 0.225W. Now, connect a 9V battery in place of the 1.5V cell. The current is now, by Ohm's law, 0.9A and the power delivered to the resistor is now 8.1W. The 9V battery must deliver far more ...

1

Ions can indeed carry current (ex. electrolysis). "An electric current is a flow of electric charge. In electric circuits this charge is often carried by moving electrons in a wire. It can also be carried by ions in an electrolyte, or by both ions and electrons such as in a plasma.[1]"

1

Well, technically yes (assuming you mean ionized and not just "charged"). Current (in simple terms) is only the time rate of charge flow, which is not exclusively limited to electrons or any specific charge carrier. Electrolytic conductivity is well documented, naturally being higher for strong electrolytes as compared to the ones that dissociate weakly in ...

1

Yes, electric current is movement of any kind of charges. The problem with your particular example is that most liquids containing ions are also conductive. Electrons will hop between molecules and equalize the ionic charges, then end up providing most of the conduction themselves. There are cases where actual ion migration results in much of the current, ...

1

The "explosion" is actually a physiological response of the body of the person electrocuted. Our muscles are basically commanded by electrical signals through the nerves to expand and contract. When large currents flow in the body from an electric shock (in the form of ion currents), these flows "hijack" the muscles, and the latter "thinks" they are being ...

1

The current required to carry a given power decrease when you increase the voltage because the power is the product of the current with the voltage (and power factor).

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