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0

I think the answer is because the book got it wrong and you got it right! Your working looks fine to me - nothing more to add.


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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 ...


0

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 ...


0

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 ...


2

Your text admits three interpretations: 1: What happens with a toy model when there's a circuit with an ideal battery and no resistance? An infinite current moves around the circuit at one moment in time (ideal batteries have a finite charge). The energy leaves the system as Electromagnetic radiation - accelerating charges radiate. Don't worry too much ...


0

From my heat transfer textbook by Cengel and Ghajar, they propose that you use the average thermal conductivity value as you specified in your last equation: $k_{avg} = k(T_{avg}) = k_0(1+\beta(T_a+T_b)/2)$ where $\beta$ and $k_0$ are material properties, assuming that the material thermal conductivity follows a linear function w.r.t temperature. You ...


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.


1

The reason it's called a voltage "drop" is because there exists an almost exact analogy between voltages and gravitational potential — that is, height. Imagine electrical current as a real, moving fluid, which is precisely how Ben Franklin viewed it (though he got the difrection of flow wrong!) The amount of this fluid flowing per second across some ...


1

Try the hosepipe analogy. Voltage is the pressure needed to push the water through the pipe. It's diameter is the resistance, and the amount of water going through it the current.


1

This is where the 0.7 comes from: Single pole low pass frequency responses are in written in the form: $$ A_v=\dfrac{1}{\sqrt{1+\left( \dfrac{\omega}{\omega_{co}}\right) ^2}}=\dfrac{1}{\sqrt{1+\left( \dfrac{f}{f_{co}}\right) ^2}} $$ for $\omega=2\pi f$ and $A_v=\left(\dfrac{v_{out}}{v_{in}}\right)$. Plots of frequency responses are call Bode plots (try ...


0

ahh, i made this very same question when learning about electricity in high-school If resistance is the slowing down of electrons and the flow of electrons is the current, why does the current stay the same when passed through a resistor? The answer (the physics teacher was good) was this: Indeed a resistor (in a sense) alters the passage of electrons ...


0

Parallel has greater current because I and R are inversely proportional and series has a greater Equivalent resistance, so parallel has a greater current.


0

The total voltage along a closed circuit without a power source such as a battery is zero. If you analyze that loop counterclockwise, there will be a drop of voltage at the resistance, that is why it has a negative sign. But the charged term has the wrong sign, because the voltage increases from right to left across the capacitor. If it is not a typo, it ...


1

What you could say is "if energy is lost in a resistor, then why doesn't the velocity of the charged particles increase, as per Work-Energy theorem?". So you're question should really go something like "Why doesnt current which is $Q/t$ increase if the velocity will increase after voltage drops"? The answer to this is that we assume all potential energy ...


0

What the resistor indeed does is to "slow down" the current. The more resistors you put the smaller the current (the current drops). If there were no resistor you would not have any voltage drop and assuming the wire does not have resistance, you will have a short circuit (maximum current flow). The circuit equations for a resistor already take into account ...


1

First, if you're going to talk about electric current, voltage, and resistance, draw real schematics, not silly stuff with pipes and pumps. This site is about physics. For plumbing go to the home improvement site. Otherwise this is a very basic application of Ohm's law. Two resistances in series will add. In parallel, their currents add. The dashed box ...


1

I mean what makes the voltage remain same across two resistors connected in parallel? To be clear, in the context of ideal circuit elements, two resistors are parallel connected if all of the voltage across one resistor is across the other resistor. This defines the parallel connection which is the dual of the series connection. If the voltages ...


0

Recall the definition of voltage: $~V= \frac{Work~done}{Unit~charge}$ Though current gets distributed, work done per charge remains the same. This is true only if, as Ross pointed out, $R_{wire}=0$, If it is not, then voltage across the resistor will be less than that across the source, because wire sections will start acting like individual resistors, and ...


0

We usually assume that wires are perfect conductors. In that case, the potential is the same at both ends of a wire, so at corresponding terminals of a parallel circuit. This is an approximation, valid when the resistors in the circuit are large compared to the wire resistance. For your last question, if the source is an ideal voltage source, switching te ...


0

In electric linear circuits theory, use is made of phasors to represent sinusoidal inputs (most common basic inputs to linear circuits). For linear circuits the total output of a given input is just the sum of each output component associated with each input component (provided there is a way to decompose any given input to compoents) (this is the ...


2

How do I identify which ones are parallel or series? If all of the current leaving one resistor enters another resistor, the two resistors are in series. The resistances of series connected resistors can be added together to find the equivalent resistance of a single resistor, e.g., $$R_{eq} = R_1 + R_2 $$ If all of the voltage across one resistor is ...


0

Since this is a homework type question, I can only offer you a tip and not solve it for you. Tip - Look at this series combination for example Notice that there is some current $I$ flowing through each of these three, and the value of potential at points 1, 2, 3 and 4 (across $R_1$, $R_2$ and $R_3$) aren't the same. On the other hand, in a parallel ...


0

You want to work this out in stages. Start at the right hand side where there are three resistors clearly in series. "In series" means that the current that flows through one resistor is equal to the current flowing through the other resistor. The simplest way to identify this is to look for nodes (points where components are connected) with just two wires: ...


0

We know that the tension between a perfect current source is 0 That's a common misconception held by those first introduced to current sources. In fact, an ideal current source produces a fixed current regardless of the voltage across. Put another way, the voltage across an ideal current source is determined by the attached circuit which is, in this ...


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 ...


1

You are right, as is Eoin's answer. I'm only answering to show one way to think about it that is useful to me when people bring up this common misconception. Imagine you have a pair of terminals with some voltage between them (like a common 1.5V battery). Nothing's connecting those two terminals together. Except for air, that is. Is there any current? No, ...


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 ...


0

What is the easiest way to solve this? The easiest way is unlikely to be the most illuminating way. What I propose is that you first work the problem from basic circuit principles and then explore more powerful solution methods such as, e.g., node voltage analysis, Thevenin equivalent circuit etc. Start from the right hand side with Ohm's law: $$I_6 ...


1

You were given the current through $R_1$ as 1mA. You should know that the 1mA current is shared between $R_2$ and $R_3$. You could determine the voltage drop across $R_1$, and then $R_{eq}$ (the equivalence of $R_2$ and $R_3$) so you can calculate the current through each resistor. Added: Yes, the voltage drop across $R_1$ is 6V. Remember that this circuit ...


1

What is the smallest resistance that can be used in a 5 V circuit if the resistors available are rated for 1/4 W? This is a trick question of sorts. It's true that one could work from the relationship $$p_R = \frac{v^2_R}{R}$$ and the range of $R$ for which $p_R \le 0.25W$. However, this completely ignores the issue of derating for reliability. ...


1

Why do we use this formula to find the total resistance? Because parallel connected conductances sum just as series connected resistances. Ohm's law is $$V_R = RI_R $$ The dual of Ohm's law is $$I_G = GV_G $$ Since, for parallel connected circuit elements, the voltage across is identical, it follows that, for parallel connected conductances ...


1

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: $$\frac {1}{R_{eq}}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ We now need ...


1

This is derived from the the formula for calculating total resistance in a parallel circuit $$\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$ taking its inverse gives: $$R_t = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}}$$


0

You have the right equation - now just solve for $R$. This means you multiply both sides by $R$, and divide both sides by $P$. In steps, you started with $$P = \frac{V^2}{R}$$ Multiply both sides by $R$: $$P\ R = \frac{V^2}{R}R = V^2$$ Now divide both sides by $P$, to get $$\frac{P}{P}R = \frac{V^2}{P}\\ R = \frac{V^2}{P}$$ Physics equations manipulate ...


1

Yes I also think that's what they did. And yes, you'd also change the voltage proportionally to the current. But all of this is as long as the resistance is constant (which is what they mean when they say the circuit is the same). A less-prone-to-mistake equation to use in these cases where the circuit is kept the same is gotten by substituting Ohms law, $V ...


1

Idealizations are just that, approximately true in many situations. If you connect a wire across a battery, the non-ideal nature of the wire and battery will show-the wire will have some resistance and the battery will not be a perfect voltage source. In most circuits, the resistance of the other circuit elements are much higher than the resistance of the ...


1

What you are missing is called conservation of charge. The fundamental fact is that electrons do not simply 'disappear' or vanish into thin air inside a resistor. This means that, if there is a certain number of electrons flowing into the resistor, these electrons must also flow out on the other side. In particular, in a steady state (i.e. the flow of ...


1

An ohm is one volt per amp. That is, a one ohm (ideal, linear) resistor requires 1 V in order for 1 A to pass through it. A 2 $\Omega$ resistor requires 2 V in order for 1 A to flow. Conversely, if 1 A is forced to flow through a 1 $\Omega$ resistor, 1 V will be developed across the resistor. Does resistance decrease the amount of electrons passing ...


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 $\hat{x}$ ...


0

When the current is turned on for 10 seconds the capacitors have gradually charged according to the equation Q(t)=CV(1-e^(-RC)) Now after 10 seconds when the switch is open the two capacitors act as two voltage sources in parallel and the current could be found out by superposition principle which is used when we have more than one source


1

After the switch is opened, we have a series circuit - the current through each circuit element is identical and equal to $i(t)$. Choosing a clockwise reference direction for the series current, KVL clockwise yields $$\frac{1}{1\mu F}\int_{-\infty}^t i(\tau) d\tau + i(t) \cdot 1k\Omega + i(t)\cdot10k\Omega + \frac{1}{1000\mu F}\int_{-\infty}^t i(\tau) ...


1

Here the capacitors look like in series but they are not actually. Capacitors can be said to be in series only if they carry same amount of charge which is not the case here.If you calculate charges will come out to be different.Look at the time constants(R*C) for the first branch and second branch, they are 1ms and 10sec respectively. As switch is on ...


5

When the switch is opened, the circuit is the equivalent of this, so I think you can clearly see that the resistors are in series and so are the capacitors. It seems that you already understand how to calculate series resistance, so I'll show how to understand series capacitance. First, you probably already know that capacitance is defined as the ratio ...


0

The best way to go about these kind of problems, is to use Kirchoff's loops. Whenever there is ambiguity about parallel/serial, Kirchoff is the way, since you disregard the question of serial vs. parallel altogether. Also remember that while resistors observe: $$ \frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2} $$ for parallel and, $$ R_T=R_1+R_2 $$ for serial, ...


0

If you cannot see that this is a Wheatstone bridge (2:4 = 3:6, so the two "arms" are balanced and there is no current in the 5 Ohm resistor) then you can use a simple analysis with just two unknowns: the voltage at the two nodes between A and B, call them x (where 2,4,5 meet) and y (3,5,6 meet). Given an imposed voltage V at A, and 0 (ground) at B, you can ...



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