New answers tagged

2

Assuming these are constant voltages, you need to solve Laplace’s equation, $$\nabla^2 \phi = 0,$$ for the potential $\phi$, with appropriate boundary conditions. That gives you the electric field, $$\mathbf{E} = -\nabla \phi,$$ from which you get the current density by Ohm’s law in differential form $$\mathbf{J} = \sigma \mathbf{E},$$ where $\sigma$ is the ...


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Actually, LC Circuit is cause of LC oscillation, When you apply kvl you get. $Q/C=Ldi/dt$, and $dQ/dt$ so after double differentiating we get $Q/C=Ld^2Q/dt^2$, which look quite same as shm equation $a=-w^2x$,so we get $w=1/(LC)^1/2$, so energy oscillate, At any instant instant energy will. Be equal. To. Field. =energy stored in capacitor +energy ...


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It is always true that the power supplied/used by a circuit element is given by $P=IV$. If you are interested in the power dissipated by a single, ohmic resistor such that $V=IR$, then either relation you ask about is sufficient. i.e. $P=I^2R$ and $P=V^2/R$ will give you the same result as $P=IV$. However, if you are looking at the power consumed by ...


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When you are calculating power consumption of some electrical device, you will find that $P=f(I,V,R)$, where you only need to know two of the three variables I, V, and R. The two known values will determine which equation is appropriate. For example: $P=IV$ $P=I^2R$ $P=\frac{V^2}{R}$


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The instant the switch is closed the current in the circuit $I$ is zero because the changing current in the circuit $\frac{dI}{dt}$ induces an emf in the inductor $L\frac {dI}{dt}$ which opposes the exactly "opposes" the voltage across the capacitor $V = \frac QC$. As time progress the voltage across the capacitor decreases and so must the the emf induced ...


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An inductor "resists" changes in the current through the energy required to build up the magnetic field. Much like a capacitor "resists" changes in voltage through the energy stored in the electric field between the plates. If you connect an inductor and a resistor in series, you will get a charge/discharge curve for the current, as with a capacitor+...


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Multimeters I’m aware of need to draw a little current from the circuit, without affecting the parameter being measured, in order to make the measurement. You don’t have a complete circuit. If you made the measurement across the terminals of one of the batteries it would give you the voltage since the circuit is completed within the battery. Hope this ...


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From your question I assume $X$ and $Y$ are named points in a circuit. What the question really means is: If you applied a voltage $U$ between the points $X$ and $Y$, what is the value of $R$ such that $U = R I$, where $I$ is the current flowing between $X$ and $Y$? So effectively it is asking to find the total resistance of all paths connecting $X$ and $Y$....


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No, you are supposed to find the effective resistance of the conducting paths between X and Y.


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You have the logical order backwards. In circuit theory power is primary and then energy is derived from that. So you start with $P=IV$ and then from that you can derive $E=\int P\ dt=\int IV \ dt$. If $V$ is constant then that works out to $QV$ but that is a derived result assuming constant $V$


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I was wondering why the current would remain the same for the portion of the wire even though the radius has decreased which would restrict flow and therefore current. Electric current $i(t)$ through a surface is defined as the rate of charge transport through that surface, or $$i(t)=\frac{dq(t)}{dt}$$ where $q(t)$ denotes instantaneous charge. If ...


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In terms of particle electrons, consider a cross section of the wire. The number of electrons passing any cross section per unit time is the same - assuming that the electrons are not pooling up anywhere. For this to occur, the density of the electrons in the thinner section has to be greater than in the thicker section. Think in terms of the flow lines of ...


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In the circuit $\mathcal E_{\rm battery}-\mathcal E_{\rm inductor} = V_{\rm resistor} \Rightarrow \mathcal E_{\rm battery} - L\dfrac{dI}{dt} =IR$ On switch on $I=0$ as the current cannot changer instantaneously. At this time $\mathcal E_{\rm battery} = L\left [\dfrac{dI}{dt}\right]_{\rm maximum}$ as $\mathcal E_{\rm inductor}$ cannot be larger than $\...


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When you say the $di/dt$ is initially "high" you are comparing it with the wrong thing, to understand what is going on. If there was no inductor and just a resistor, the initial value of $di/dt$ would be (in theory) "infinite" as the current changed instantaneously from $i = 0$ to $i = V/R$. The "back EMF" $e_L$ is initially equal and opposite to the ...


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Assuming you are talking about a series RL circuit with an ideal inductor, it is correct that $\frac{di}{dt}$ is maximum and the voltage across the inductor is a maximum equal to the battery voltage when the battery is first connected to the circuit. The current is initially zero. It is also correct that the rate of change in current then decreases. However,...


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As @Steeven points out, the battery polarity is actually the reverse of that shown. Just like loop currents in loop analysis are generally initially unknown, the center battery emf here is an unknown. For loop currents a direction is initially assumed. If after solving the loop equations a loop current turns out to be negative, it simply meant the assumed ...


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This is a "homework and exercise" problem that we generally don't provide solutions to, but only guidance. So here is some guidance for you to do additional research. If I remember my filter concepts correctly (its been a while), this appears to be what is called a low pass filter, i.e., a filter that permits low frequency signals at the input (left side) ...


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For calculations, it depends on the features you want to model. This won't happen in a physical circuit. If you don't want to get to fine detail and start worrying about propagation delays, then the simplest way to avoid it is to look at the inductance. Any real circuit will have a non-zero inductance. You can think of this as the circuit's mass or ...


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If there is no voltage change, then how would the capacitor become charged? I don't follow your logic here. It's true that there is a voltage drop across a resistor if there is non-zero current through the resistor. If there is zero volts across the capacitor (at some time), then all of the battery voltage is dropped across the resistor (if you don't see ...


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In fact, it is the voltage across the capacitor that stops the charging process. There is no problem running current through the capacitor when the voltage across it is zero. When the switch is thrown (not shown), there is a potential difference between the battery terminal and the capacitor. Charges will try to accumulate on the capacitor plate. ...


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Contrary to what the other answers say, resistors do change their value with time, even the most accurate ones, even at perfectly constant temperature. This is due to various phenomena, e.g. release of internal stresses, contamination from impurities etc. For instance, National Metrology Institutes keep historical records of the drift of their standard ...


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There is no SI unit for the time derivative of resistance. That is probably because resistance does not normally vary as a function of time.


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Resistors are usually a constant value. They may vary with temperature but you'd need to know the material the resistor is composed of. If you just search on resistors you should be able to find technical sheets for different resistors that will have information as I described. The resistors should be constant in time so $d\Omega/dt$ = 0.


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This is a very good question and you are not alone in finding it difficult to decide which resistors are in series and which resistors are in parallel. So go back one step and think about the reason that you want to know if resistors are in parallel or in series. The reason is that if you can identify series and parallel resistors you have formulae which ...


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"If the current must travel down two or more paths, items on those paths are in parallel until those paths reunite" I don't believe this is generally true. Consider the case of one path being a resistor and the other path being two series connected resistors. For this case, there are no parallel connected resistors. "If two resistors share the same ...


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Are these all legitimate ways to work determine whether resistors are in series parallel The first and third, with the exception of some slight tweaking, are the most common versions I have seen and seem legitimate to me. The other two are, at least to me, problematic, for the following reasons. "If the current must travel down two or more paths, items ...


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This maybe of some help. I feel the question is missing something, if you could provide more clarity as to what you are looking for, I can give you some insight.


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Your first equation is correct for $v_{input}$ making a current through a series resistor $R$ into the feedback input of the opamp. Your second equation is for a capacitance $C_{detector}$ across the opamp inputs and is incorrect because the voltage $v_{input}$ on $C_{detector}$ at time $t$ is not $\frac {q(t)}{C_{detector}}$. Instead, the opamp has worked ...


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Proportionality requires a constant. $R$ is a constant (it doesn't change when $V$ or $I$ change). $W$ is not a constant (it depends on $I$; higher current causes more work done).


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What exactly is happening above? Note that the $R$ in $$V = RI$$ is ordinarily understood to be a constant. We say something like For the ideal resistor, the voltage $V$ is proportional to the current $I$ where the constant of proportionality is the resistance $R$ In your post, you wrote an equation involving the work $W$ $$V = \frac{W}{T}\frac{1}...


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In the relation V=W/IT potential I s not exactly inversely proportional to I because if we increase current I then the work done W will also increase by I^2 as W=I^2RT. So the W will also increase and the value of potential will also Increase. V=W/IT Let for T and R be 1. Then V=I^2 x 1 x 1/ I x 1 V=I^2/I If we increase I from 1 to 2 then potential will ...


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If $V$ would be inveresly proportional to $I$, then there would be a constant $k$ such that $$V = \frac k I$$ Sorry, in your formula $$V=\frac W{I T} $$ which is the same as $$V=\frac{\frac W T} I$$ the part $\frac W T$ is not constant. $\left(\frac W T \right.$ is work by unit time, i. e. power, which is not a constant independent from $V$ and $I$.)...


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Ohms law gives the fundamental relationship between voltage and current for a resistor. V is proportional to I where the proportionality constant is R. The other equations are manipulations of various combinations of ohms law and power (P) or work (W). When the various equations are manipulated, the constants of proportionality change and contain other ...


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The reason that lightning strikes a lightning rod rather than another point is that the lightning rod provides a path of least resistance from it to the ground. This is the reason for the 25 Ω limit in the regulations. However, if you want to use electricity to heat something up, you need have a high enough resistance in the path of the electricity to ...


1

Lightning can boil water. The reason why many objects explode when struck is that the water they contain vaporises. So there is enough energy available. However, lightning is a very transient phenomenon, so the amount of water you could manage to bring to boiling point, without wasting heat on steam explosions, would be dependent on passing the heat energy ...


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The real question here is "how much water"? But let's start with the literal question. Heating water to the boiling point (373K) just means that the lightning current heats up a resistor in the path to 373K. That means the resistor shouldn't melt at those temperatures. Is that physically possible? Sure - plenty of metals will melt at much higher ...


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You are forgetting that $W$ is not an independent constant. In fact $W=I^2 R\ T$ so that $\frac{W}{I\ T}$ is indeed not inversely proportional to $I$.


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For most (finite) DC circuits, you count the number of holes in the circuit, (six in the case of the cube), define a current around each hole, and then write a voltage loop equation for each. This gives you a number of simultaneous linear equations. If you are allowed to use a laptop computer, a good spreadsheet can solve these using matrix operations. (I ...


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If you know the formula for capacitance of two capacitors in parallel capacitance for two capacitors in series then you can do a simple step by step simplification of the diagram as below. I haven't shown all the steps but start with calculating the capacitance of the two caps in series at the top left of the rectangle (= capacitance shown in blue), then ...


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The inductance of the inductor, $L$, and the capacitance of the capacitor, $C$, control the rate at which energy oscillates between them. The period of oscillation is $2\pi\sqrt LC$ which means that as the magnitude of the inductance gets larger then the period of oscillation gets larger which can be interpreted as a larger inductance resisting the change in ...


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For a simple resistor (in which we can ignore possible thermal effects of increasing current), if you double the current then the voltage also doubles - while the resistance remains constant. So R = V/I and R = (2V)/(2I)


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Resistance is a propety of electrical components, not a consequence of the active circuit - a 100-ohm resistor has a resistance of 100 ohms, whether it's hooked up to a AAA battery or the power grid. In Ohm's Law, the quantity $R$ is typically fixed. So, when you double the current running through an element, the voltage across that element also doubles, ...


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If we have a given physical resistor, and we want to double the current through it, we do that by doubling the voltage across it. The resistance, at least ideally, stays the same. In the real world, the resistor value will change somewhat due to the temperature of the part rising. Maybe a few 10's of ppm for a high quality resistor, or several per cent if ...


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The noise observed at a contact is known as chattering. A stronger mechanical contact should impart a steadier electrical contact resistance (ECR) [1]. However the structure and cleanliness of the surface should also be considered in the design of this setup to minimise the presence of varying passivating layers. It should be noted that time further plays a ...


2

Indeed the contact resistance can be, as discussed in previous answers, attributed to surface features in terms of asperities and passivating layers. The behaviour of these barriers to conduction depends on contact pressure. Passivating layers are oxides and hydroxides that ubiquitously form on conductor surfaces, and pose a barrier to electron transport. ...


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"if you apply an increasing voltage to one plate this will force charged carriers to that plate" I disagree with this. The negative charge carriers will be attracted toward this positive plate but will not move there unless the capacitor fails. The electrons will pack on the plate plugged to the negative terminal, i.e. where the electrons originates from. ...


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If during self-induction the inductor produces a current in opposite direction of that of battery in order to resist the change of current through it then how could the current after some reasonable amount of time gets to its peek point? If you apply a DC voltage to an ideal inductor, it will never reach its peak current. The current will continue to ...


1

The whole point is that for a normal conductor Lenz can never win. It is true that at the start there is no current but there is a finite rate of change of current with time and so with the passage of time the current will change from its initial zero value. Another way of looking at the situation is that if Lenz did stop the change of current, and ...


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As you will see from the circuit diagram the resistors are used in a variation of the Wheatstone bridge arrangement. With such an arrangement a balance point, zero current through the galvanometer, needs to be found. This is achieved by tapping a jockey, spade ended conducting contact, along a uniform resistance wire, $A$, until the zero current position is ...


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Addition of salt will facilitate the conduction of electricity due to ionization. The extra ${Na^+}$ and ${Cl^-}$ ions will increase the conductivity of the medium. 1)Voltage will be slightly affected if you are considering internal resistance. 2) and 3)Current will increase due to the decrease in resistance in the conduction medium.


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