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I think that your reasoning is correct: A standard circuit breaker should not trip if the resistance of a circuit is increased. However, in recent years there is a new kind of house circuit breaker called an AFCI circuit breaker which may trip if there is a loose or poor connection in a circuit. That may be the type of circuit breaker that they were talking ...

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If the circuit used alternate current, it might happen that a wrong terminal would introduce additional series capacitance or inductance; these could cause the breaker trip if the original load was mostly inductive or capacitive respectively. But it is not probable that the effect of a termination would be sufficient for this. Instead I conjecture that ...

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You set up a circuit which contains the galvanometer and adjust the circuit so that the deflection on the galvanometer is a maximum (for greatest accuracy). A calibrated resistance box is connected in parallel and adjusted until the deflection on the galvanometer is half of the deflection with no resistance box. In this condition the resistance of the ...

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In such circuit see that all resistance at rim have same potential, since no resistance reduced it. At rim all have same potential as that at $a$. At center all have same potential as that at $b$ So, all have same potential difference $V_{a} - V_{b}$ Same potential difference means that they are connected in parallel.

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I think that you have faced a conceptual problem because you have used the relationship $V=RI$, rather you should have been using the definition of resistance $R = \frac V I$. Then as $V$ goes up and $I$ goes down the value of $R$ increases. Also you must be careful about $I=0$ (current equals exactly zero) and $I\rightarrow0$ (current gets closer and closer ...

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Bear in mind that infinite resistance is equal to an open circuit. In real life if you have a battery not connected to anything, there will always be a voltage across the terminals. Even a lead-acid car battery will produce sparks if you short the terminals!

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It depends on the shape of the medium, but basically you could have an infinitesimal resistance, $\mathrm{d}R$, and integrate that. For example, for a uniform area wire of area $A$ and length $\ell$ in which the resistivity varied sinusoidally about some nominal value, $$\rho(x)=\rho_0(1+0.05\sin\left(\frac{5}{\ell}x\right)),$$ then $$R = ... -1 At R→∞ voltage won't be 0 but little less than EMF off cell, since a series circuit with voltmeter and battery is complete. Yes, voltage across resistor reaches 0.Voltmeter reading is not EMF because it draw some current which pass through internal r and gets some potential across it. An Ideal voltmeter cannot tell you voltage across resistance in ... 0 -First, I'm trying to grasp the concept of why and how the voltage drop at the terminals of a battery depends on the resistance of the circuit. The EMF, \mathcal{E}, exists chemically and is always present. If there is a path for current to flow, then there will be a voltage drop across the internal resistance, r, and the external resistance, R. ... 0 Assuming steady state means that you have no current flowing in the branch with capacitors in it. So think of the circuit first as a potential divider made up of the battery and the two resistors. From this you can find the potential difference across the two capacitors. 2 You have two basic options: Realize that this is actually just two resistors in series with three parallel resistors, and analyze it using the equivalent resistances of resistors in parallel and series, or Use Kirchoff's laws to derive the equivalent resistance. If you choose option 1, I'll help you out by revealing the parallel resistors in this ... 0 You are using the word "resultant" in place of a more precise mathematical term - "net", or "sum". Add up all the forces on an object, and if they equal zero, the object experiences no acceleration. For your box, since the forces are in opposite directions,$$\sum F_i=F-F=0$$For the car, the forces are actually quite complicated, but start with the fact ... 0 In practical terms, resistors in electrical circuits can be used for a wide variety of applications. If we put a small resistance in series with a conductor in a circuit, and know the value of the resistance accurately then we can measure the current through that conductor by reading the voltage drop across the resistor and using Ohms law to calculate the ... 1 Short answer: "neither". A resistance, in a circuit, causes a voltage drop for a given current. Alternatively, if the voltage is given, it causes a particular current to flow. As soon as a circuit gets a little bit more complex (more than one battery-resistor-light bulb-switch), it becomes completely impractical to use lots of batteries to control ... 2 An electromagnet has a lot of turns of wire, which gives it a property called inductance : the property of resisting a change in the current. You may be familiar with Lenz's Law$$V=-L\frac{d\Phi}{dt}$$which says that an inductor with inductance L will generate a back e.m.f. V when the flux \Phi through it changes. Now if you drive a current through ... -1 Google translate is a wonderful thing: Cut and paste the solution (in parts) into Google translate. It will not be a perfect translation but it should be good enough to provide an explanation as to how to solve the puzzle. For example here is the first paragraph: 20th edition , the role of VI . 3 ... čtverákčtverec ( 4 points , average 2.73 ; dealt ... 0 This is a very interesting question which has at least two answers. The first answer is for an electrical/electronic engineer who wants to get a correct answer by using any systematic and coherent method. The Associated Variables Convention is used. You assign a current direction to each circuit element and then assign a plus sign at the point at which ... 1 Assuming the bulb emits in the visible wavelength regime, then you can use a silicon photodiode. Depending upon the particulars of your "bulb" you could put the photodiode in series with a resistor and put that combination in parallel with the bulb. So as the photodiode receives more light, it's resistance will decrease thereby allowing current to flow in ... 0 emf induced by inductor is negative but emf applied by battery on inductor is positive. So, that inductor tries net voltage across it is 0. Emf applied by battery is E°sinwt of which LdI/dt is applied across inductor, RI across resistance and Q/C across capacitor. -LdI/dt is induced across inductor and not applied. Now, consider direction of emf is ... 0 When you find the response of a linear circuit for a frequency \omega you are taking the differential equations that describe it and looking for the homogeneous solution. The true response of the circuit is this homogeneous solution plus a particular solution which depends on the initial conditions. That is the missing term you found. The particular ... 0 Duct tape should be okay. A few small alligator clips from the hardware store would be better. I assume you're going to twist the wire around a solid thin probe. That connection point will introduce uncertainty in your result because of the uncertainty in length and contact resistance. The contact resistance will depend on the surface area contact which ... 1 The electron moving from negative to positive terminal has to suffer lot of collisions with other electrons and has to also interact with the kernels present.Hence greater the length,, greater is resistance. 0 If you ask this question in full generality -- i.e., is it possible for any value of R1 to R9 to infer their value from the resistances between A..F -- then answer is no. Just imagine R_1=R_2=R_4=0. You will never be able to infer R_5, as there is a short circuit. Regardless of that, I doubt that there is generically a unique solution, since the ... 0 By connecting A with B and E with F you can reduce the problem to a 2x2 matrix. Same for all other combinations of two (or n-1 lines in general). I would then call the resistors R_{11} to R_{22}. If you short A with B and E with F, you combine resistors in such a way that R_{22}=R_9. Shorting out a horizontal line with a vertical line in the simplified ... 0 Is this part of a homework problem? Low resistance - think about how you would connect the resistors to the bridge. High resistance - think about how you note when the bridge is balanced. The answer to your question is in all three Physics textbooks that I have opened as well as a couple of webpages. 0 Input resistance and output resistance are parameters to help you use a particular circuit. In your example above the attenuator starts at Port 1 and finishes at Port 2. Removing the load resistor R_L for the moment suppose you were asked what the input voltage to the attenuator V_i was with the given supply. To answer such question you could reduce ... 0 I know that input resistance is the resistance as seen by the input terminals As written, this isn't correct. The input resistance is the resistance looking into the input terminals. Conceptually, this means that if one changes the voltage across the input terminals (only), the input current changes by$$\Delta i_{i} = \frac{\Delta v_{i}}{R_{i}} ...

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You don't need to supply any extra energy to return electron to + ve plate . Electron already have enough KE even after passing all the acquired energy by battery to resistor. Never forget that actual speed of elections is very huge. When these electrons were in battery they already had enough KE. When you applied resistor it just increased its energy. ...

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The currents will be the same. Using Kirchoff's current law, the total current going out of any node or any closed contour is 0. Suppose you have two resistances connected in parallel between nodes A and B. Draw a circle around the parallel resistances. The total current going out of the circle is 0. So outgoing current at A + outgoing current at B = 0. In ...

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Looking for something else accidentally found your conversation Probably is useless after 3 years but let me make a comment Nothing new, but putting it in on other way Keeping the current fix in the formula Q=I2Rt, shows that the energy amount converted to heat in a resistor is proportional to the resistance. This is because in order to keep the current ...

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In order to make the problem symmetric, consider this: What happens, when you take the dodecahedron and drive current $I$ into it from vertex $A$ and drive $I/20$ out from all every vertex (including $A$)? By Kirchhoff's laws and symmetry, there is a current $I_{A-out} = \frac{(I-I/20)}{3} = \frac{19}{60}I$ going from $A$ to neighbouring vertexes. Now ...

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So the spectral approach as I remember is you make an anszats $i=\sum_k a_k f_k(t)$ or something like that, you use some theorem to say that each component of the sum is independent so you get $a_k f_k''(t) + 5 a_k f_k'(t) + 1/4 a_k f_k(t)=0$, to get the factors (characteristic polynomial is what you said I believe) My guesses where the error could be: I ...

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I had read that resistances only lose energy in the form of heat. Correct. Is that what the authors are talking about? Yes, the author is just using confusing language. Do they mean to say that this internal energy is first stored in the resistor in the form of internal energy and then dissipated as heat? The author means that electrical ...

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The potential difference across them wouldn't be equal because their lengths are equal. It would only be equal if their resistances were equal, which as you point out they are not.

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If they are in series, then yes the current will be the same flowing through each, it has no other path to take.

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