# Tag Info

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Dissolving salt in the water creates sodium and chloride ions which in the presence of the potential of the battery provide a path for current flow, the movement of charge. Thus resistance is decreased and current is increased. While an ideal voltage source would see no decrease in the voltage, a real world battery has its own internal resistance, and so ...

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Without even doing any circuit calculations, you can conclude the voltage between a and b is zero by symmetry. Proof: Assume there's a voltage between the two points. If you close the switch, a current would flow. If you take the mirror image of the circuit, you'd expect the same current to flow, but in the opposite direction. Except the circuit is left ...

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First, let's assume the left-most terminal is connected to the positive terminal of the battery and the capacitor voltage reference direction is left-most terminal positive. Now, consider a KVL loop clockwise through the top 2C capacitor, the switch, the bottom C capacitor and the battery: $$10 \mathrm V = V_{2C_{top}} + V_{ab} + V_{C_{bot}}$$ So, the ...

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The potential difference across the top two capacitors must be the same as the difference across the bottom two. I will number the capacitors $C_{11}$ for top left, $C_{12}$ for top right etc. If we assume the charge on each capacitor is the same, then the voltage difference must be zero. But if we can assume that each capacitor may have a different charge ...

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If the question is viewed correctly the batteries are not connected to each other, therefore it is an open circuit. Or if it is connected the batteries are facing each other and the values are not mentioned and so let us assume they are equal. Then the net voltage is zero which is the same as no voltage so V across the 4 ohm resistance is zero.

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Are you asking why the electrons in wires exhibit this behavior? If so it is due to the fact the path of moving electrons is curved when in a magnetic field and how much it curves increases when the magnetic field increases. Since the electrons in wires are always moving about randomly and never still, their paths' get curved. This net rotation of electrons ...

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Expanding on Kavan's answer, I'd say that since the two ends of the cell are facing each other (as in they cancel out) that the dots could also be interpreted to do the same. Usually dots correspond to a series that follows the same pattern given by the first term (in this case the cells that cancel out). So I'd say 0 is your best shot. If the cells aren't ...

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I think that, if both batteries have equal potential difference, then the net potential difference will zero between any two points of circuit, so the answer is zero. However, this is a guess, since no information is given and both batteries seem identical in the figure.

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$V = iR$ but $P = iR^2$, so if the current, $i$, stays constant but if the resistance, $R$, increases, then the power, $P$, increases too.

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If you are comparing two voltages with identical currents, you cannot be talking about the same bulb in both cases. This means that you are comparing two different bulbs, and there is no way to tell which will be brighter, since different bulbs can be designed for different luminous efficacy, which is light per unit power. For instance, a bulb can be ...

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Yes, if you go around N times then the emf will be N times greater than if you went around once. Why? Suppose one loop bounds a surface with area A, then N loops will bound a surface with area perpendicular to the field NA. Imagine you're looking directly through the solenoid and the surface it bounds is made out coloured glass, how many layers of glass are ...

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Q1. "from the above values we can calculate Power(P) as P=V∗I" A1. Yes. Power=10W Q2. "If voltage is amplified or raised to 4 times that makes V value to 20 V, what happens to the values of current and power." A2. Assuming your load is a resistor, then your original load resistance was 5V/2A=2.5Ω. Therefore, if you increased the voltage to 20V, your ...

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A resistor is defined as the circuit element for which the voltage across is proportional to the current through and the constant of proportionality is the resistance $R$: $$V_R = R\cdot I_R$$ Clearly, for this linear relationship, it is also true that $$\frac{dV_R}{dI_R} = R$$ However, for general circuit elements, the derivative of $V(I)$ is not a ...

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$R(V,I) = \frac{V}{I}$ by definition, it is not a gradient. $r = \frac{dV}{dI}$ is called the fractional, differential, dynamical or small-signal resistance. It just happens that for resistors $R(V,I) = R_0$ is a constant, thus the two quantities are the same: $r = R_0$.

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My argument was that because the resistance is higher, there must be less voltage going through at that point. This is probably the cause of the confusion. In spite of the usual formulation $V=IR$, in an electrical circuit Voltage and Resistance are the "inputs" to the equation and Current is the result or output. As an analogy, think of Newton's 2nd ...

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The problem is really with the parallel diagram on the far left. It shows a 6V drop across the parallel combination of resisters. That is determined backwards from how we read left-to-right. In order to solve for the voltage drop you must: (1) Solve for the equivalent resister to the pair, the 20 Ohm resister shown in the middle diagram. The equivalent ...

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You are right. Voltage is an electric potential difference. The concept of potentials is more general (e.g. gravitational potential) in physics.

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Everything you say is correct in the steady state. The problem you run into is that when you remove charge from a charged capacitor to an uncharged capacitor, there is a potential difference. And somehow, you have to remove the energy from the electron that moves from one to the other. It turns out, as you calculated, that you in fact remove half of the ...

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If you take the hosepipe analogy, the total amount of water flowing through the pipe is charge, the amount per second is current, resistance is resistance and in order to get a flow though the resistance you need pressure. Which is the voltage.

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Yes. The suit is of a conducting material and charge thus moves inside the suit material much more likely than through the content (the person) inside the suit.. This is called a Faraday cage and is an effective shielding mechanism from electric shocks. Excess charge will stay on the outer surface of a conducting object and charge will redistribute on the ...

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Electrons (charge carriers in a wire) move from high electric potential (high voltage) to low electric potential(low voltage). While electrons are travelling, it is the resistors which pick the amount of electrical energy they want (per their electrical capacity) and it is not the electrons that determine how much they should drop off at each of the ...

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The question is ill-posed; the electrons "know" nothing, and voltage is not a property of the electron (other than e.g. charge, which is a property). In fact, voltage is a pretty abstract concept; it is energy divided by charge. And that means explaining an abstract term by another abstract term. Let's be more fundamental: nature shows that charges exert ...

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BEWARE THE SANDWICHES!!! :) In the spirit of math-avoidance sandwich-juggling, here's a better analogy, a visible one. The movable charges within conductive circuits are like silver bead-chains, like those little chains which attach the pens to desks in old-school banks. (Growing up I always played with these when mom was in the teller line. Do those ...

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Just a simple answer - there's nothing to dissipate energy. If there were a resistor in the circuit, it would dissipate energy as heat. Inductors and capacitors don't dissipate energy. The energy just sloshes back and forth between being stored in the magnetic field, and being stored in the electric field. It's just like a spring-mass system, where energy ...

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Wouldn't this inductor's emf counteract the discharging capacitor and actually charge it? / stop the capacitor from fully discharging? The inductor doesn't care about what the charge state of the capacitor is. All it cares about is how quickly the current through it is changing, and it generates a back-voltage according to the equation V=L*dI/dt. You ...

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The inductor never creates a current in the opposite direction. An inductor creates an EMF to counteract the changing B field(Lenz law). The B field is changing because the current in the inductor is changing. So effectively, the inductor resists changes in current. So initially, the capacitor tries to discharge strongly but is slowed down by the ...

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Electrons move because they are in a region of space with a non-zero electric field. They don't accelerate to high speed in a wire because they keep bumping into things; a kind of friction which dissipates energy much like the friction you are used to that explains why resistors get hot. In effect their speed depends on the strength of the local electric ...

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There is an error in your multiplication. $$(9 \times 10^{9})(20 \times 10^{-9}) \neq 90\times20$$ $$(9 \times 10^{9})(20 \times 10^{-9}) = (9\times20)\times10^{0} = 9\times20 =180$$ This means that your answer is off by a factor of 10. Without the error your answer should be 3kV.

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All else the same, KV is inversely proportional to the stall torque of the motor. But, all else is seldom the same, and stall torque often isn't that useful a metric.

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For a linear DC motor the "back emf constant" in units of volts/rad/sec is equivalent to the torque constant of the motor in units of N-m/Amp. But the manufacturers will usually specify the back emf constant in units of volts/RPM, so the numbers will differ. Strictly speaking a brushless motor is no a linear DC motor, but rather a synchronous motor, and the ...

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There is another constant used to describe electrical motors that quantifies the torque, the torque constant Kt $$K_t =\frac{\tau}{I}$$ where $\tau$ is the torque, and I is the current. Therefore, the torque a motor produces is $\tau = K_t I$. The SI unit of Kt is the weber. It turns out that: $$K_t = \frac{1}{K_v}$$ So a lower Kv implies a ...

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