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0

I think you understand that the series has more resistance than the parallel, so more current should flow in the parallel case. Since there is more current in the parallel case, the battery has to supply more current so it is more stressed, and it gives out a lower voltage. So its $V$ (notice that $V$ is really just measuring the voltage drop across the ...


0

There is a 90 degree phase difference between the input current and the induced voltage in the ring because of Lenz's law. Then, there is a 90 degree phase difference between this induced voltage and the induced current because of the self-inductance of the ring. This amounts to a 180 degree phase difference between the input current and the induced current ...


2

I think you might consider first special relativity. We can model the problem as being in Minkowski spacetime with cartessian coordinates and put the switch at $(0,0)$ and the lamp in the coordinate $(0,L)$. Where $L$ is the distance from the switch to the lamp. Then the question is what is the state of the lamp at $(120,L)$? Using a spacetime diagram ...


3

If you count the number of times the switch is flicked, then when the number is even, the lamp is off, and when it's odd, the lamp is on. So we can rephrasing your question: is infinity even or odd? That's one for mathematicians... they will probably say "both". So the short answer is - there is no "real" answer to your question. But most likely the lamp ...


3

The "derivation" you describe is valid at a particular moment in time. $$\begin{align}\Delta E &= V \Delta Q\\ \frac{dE}{dt} &= V \frac{dQ}{dt} \\&= VI\\ P(t) &= V(t) I(t)\end{align}$$ I added the dependence on time explicitly. To address your comments: $W=QV$ is only true when $V$ is constant; you can't simply take the derivative and ...


3

Power is an instantaneous concept. $P=IV$ gives the instantaneous power at a given instant of time, given $V$ and $I$ at that time.


1

The exact equations for I-V characteristics of transistors are derived using quantum-mechanics. Several approximations can be used, one of which is based on the shottky barrier analysis This reference here derives the I-V linear and quadratic approximation (in saturation) for FET transistors. Another reference here UPDATE: As @QMechanic pointed, ...


-1

They should because light bulbs connected in parallel definitely give you more energy per unit time :). It is because in parallel you have to circuits...for example, if you have a simple circuit with one bulb, and if you have total of 4 A of current, what will the current be if you add one more, in parallel? If you remember, sum of currents in two branches ...


0

Situation 1., the ideal wire with an ideal voltage source, is an idealization: it is not a situation that can occur in nature. One should not be surprised that physics, which aims to describe nature, cannot describe a situation that cannot occur in nature. I will note that you specified a battery, not an ideal voltage source. In this case one can develop ...


0

According to http://www.electronicspoint.com/threads/bump-n-go-robot-help.64701/ : "These kind of toys usually have a simple mechanism driven by a single motor, it's not easy to describe tho. Usually, a vertical motor drives a pair of wheels in a free turning cage. It's a kind of differential - imagine the solid back axle of an old car with the prop shaft ...


1

I think that the 3 mA flows through the 3k resistor only - because that gets me to the answer you gave. The approach you need to take is this. The same voltage (3 mA x 3 k = 9 V) exists across the 3k and 6k resistors. You know the current through the 3k, so you know the current through the 6k (9/6k = 1.5 mA). Then you know the current through the 2k (1.5 + ...


1

First, for simplicity, assume the diode is ideal. For an ideal diode, the voltage across cannot be positive (the voltage at the anode is either equal to or less than the voltage at the cathode). Since the cathode of the diode is connected to the 0V reference, and since the anode is connected to the output node, it follows that the output voltage cannot be ...


0

For simplicity, I'm going to assume a 0.7V diode drop in the ON state. In your diagram, you can call the low node ground (0V), in that case, you have one of two situations: The diode is on (closed switch) In this state, current can flow through the resistor and therefore, through the diode. In this case, 0.7V is dropped across the diode and the remaining ...


0

in P=VI so when connect the the batteries in parallel then voltage is improving and its current increases because of the less of the resistance ana powrer is directle proportional to both thats why power is also increases....


0

The diode has a central barrier at the PN junction which allows charge to flow across it only when a sufficiently positive voltage is applied to the P part with respect to the N part so that the barrier potential is overcome.In this case the diode is said to be forward biassed and conducts current.At all times when the P part has a voltage less than the ...


0

From the information you have put in your question I think that in the reverse bias case the 'diode is open' means that it behaves like an 'open circuit' or 'open switch' and no current passes through it and thus no current goes through the battery. The load resistor has current passing through it that goes to circuit/components not shown on the right hand ...


0

It depends on what is meant by "added to the lamp". If the resistor is in series with the lamp, then there will be a voltage drop across that resistor (since a current is flowing) and less of the potential is available for the lamp. But if the lamp is modified so it has higher resistance (a perfectly acceptable interpretation of the wording) then the lamp is ...


0

If you add a resistors in series to another resistor (lamp), the voltage across the lamp will decrease. The new voltage will be: $V_{LAMP}=R_{lamp}V_{total}/(R+R_{lamp})$ The resistance in the lamp is a constant property. Regardless of the change the resistance is constant, it is the voltage across it that's different.


0

Look, for example the page at http://farside.ph.utexas.edu/teaching/302l/lectures/node57.html . It seems that the EMF E is a characteristic of the cell. So it should be constant, whatever hapens that is non-destructive. Regarding the second question, both formulae are correct, and both refer to the same voltage V between the points A and B in the figure of ...


0

Phasors are a way to analyse linear circuits using fourier analusis for the input, so the output can be analysed per frequency component (definition of linear circuit). (see related answer) Any (pure) DC component appears as a constant term (or bias term). Each phasor represents only one frequency component $I(\omega)$ But one can also take limits when ...


-1

First of all, as the comments on the question point out, the current is infinite through a wire only if the entire wire is an ideal one. Meaning, the resistance of the entire wire is zero; then, not rigorously speaking, Ohm's law becomes I = V/0 resulting in an infinite current. If there is a simple ideal wire (a single loop) with a resistance R at some ...


0

The voltage that drives the current is different in different parts of the circuit. The sum of the voltages driving current through each part of the circuit, wire and resistor must be equal to the voltage applied with a power source or battery. In the case of the circuit with the resistor and wire then nearly all the voltage will be pushing electrons ...


0

Crudely, electrons repel each other and even out the charge. While the influence of the electrons travels at a good fraction of the speed of light the electrons themselves do not move much. From this link " The electricity that is conducted through copper wires in your home consists of moving electrons. The protons and neutrons of the copper atoms do not ...


0

You're allowed to be confused on this issue - for many students it's a new phenomenon for which no analogy is perfect. Current is very easy to understand. As you mentioned it's basically just electrons flowing past per second. However, as it would be rather time-consuming always saying things like "the current is 1000000000000000000 electrons per second", ...


1

my main question is, if the energy is not the kinetic energy of the electrons, what does in fact bring energy to the resistor and how does it heat up? We assume steady state operation. The drift velocity of the electrons entering the resistor must equal the drift velocity of the electrons leaving the resistor. This follows from the fact that the ...


1

Yes, it is the electron kinetic energy gained between collisions with the lattice atoms which is transferred to the lattice atoms who heats the lattice. After collision the electron changes it direction essentially, so the drift velocity is small whereas the instant velocity is high. The drift velocity has nothing to do with the temperature. In absence of ...


2

In addition to the good answers above here is something to think about. If the resistance is zero then for a current to flow there does not need to be a battery - the emf can be zero. No work is done as the current flows. This may sound like a strange case, but it is how many strong magnets work in nmr (nuclear magnetic resonance) spectrometers. The have a ...


6

Your original text admitted three interpretations, and I'm leaving the answers here: 1: What happens with a toy model when there's a circuit with an ideal battery and no resistance? All the charge moves around the circuit at one moment in time (infinite current). The energy must leave the system as Electromagnetic radiation - accelerating charges radiate, ...


1

It is very simple, to pass current, it needs a path to follow unlike voltage (potential) which you can have with an open circuit. Think of it this way, electrons need to jump from atom to atom, this will happen until it reaches an end to the wire or conductor. At the end of the wire that ends, the electron stop and build-up to the highest voltage it can ...


1

Why is it so? Well, it isn't actually always so. It will depend on the actual circuit configuration and whether the switch opens or closes when $t=0$. But first, here are a couple of crucial results to always keep in mind when solving these type of switched circuits: the current through an inductor must be continuous the voltage across a capacitor ...


1

@Floris' answer being good, i'll give another view on the matter. A capacitor is equivalent to an open circuit (since simply put, a capacitor is an element consisting of two plates which do not actually touch but through another medium, the dielectric, the circuit is not connected at that point where the capacitor is located), whereas an inductor is ...


1

Circuits with inductors are sensitive to changes in the signal - think of them as differentiators. Circuits with capacitors are responding to the integral of the signal over time. When you first turn on a circuit, the current wants to make a step change - which the capacitor doesn't care about, but the inductor resists vigorously. Thus the current will flow ...


0

The beginning statement is incorrect or, at best, misleading. There could, in fact, be a current through the ammeter and it would still be the case the potential at $C$ equals the potential at $D$. Let me explain. The voltage across an (ideal) ammeter is zero. Thus, with the ammeter in the circuit as pictured, it is given that $V_C = V_D$ - the presence ...


0

Ok so this is the Wheatstone bridge (see http://en.wikipedia.org/wiki/Wheatstone_bridge). Imagine that in the top diagram the $V$ is $10~V$ and all the resitances of the resistors are the equal to each other - say $100 ~\Omega$ each. Let us say the potential at E and F is $0 ~V$, and at A and B is $10~ V$ - because the resistances are all equal the ...


1

"a circuit that has different resistors at extremely different temperatures" -- Each resistor independently puts out its own noise related to its own temperature. "one long resistive element that has a temperature gradient across the whole thing" -- That's actually the same thing again. Treat it as a large number N of resistors in series, each with ...


1

Simply put, does the voltage drop on each resistor mean that there are more electrons on one side of the resistor compared to the other side? Yes, the charge density at one end of the resistor must differ from the other if there is a current through. Consider, for simplicity, a resistive element of length $L$, area $A$ and resistivity $\rho_r$. ...


0

The equivalent resistance of a network is that single resistor that could replace the entire network in such a way that for a certain applied voltage V you get the same current I as you were getting for a network. If your network is "a bunch of resistors in a box", and you can't look inside the box (you just have access to the two terminals sticking out), ...


0

The resistance of two parallel resistors is defined by: $1/R_e = \sum1/R_i$ For two series resistors, it's simply the sum of the individual resistances of each resistor.



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