# Tag Info

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Because of magnetic field attraction towards the coil.

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You have two sources of error: error in the source voltage, and error in the current measurement. Since your graph is log-log, at the bottom left corner (small currents and voltages) any errors are greatly magnified; even a 50mA or 100mv error looks big. At the top right corner, however, the (probably of a similar magnitude) errors are swamped by the signal. ...

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At very low voltages (and currents) the power produced is small, so the filament doesn't produce much heat. At these low temperatures the bulb tends to have significant interactions with the surroundings, especially drafts which will cool the bulb in an unpredictable fashion. Another source of error may be the bulb contacts, which may be showing significant ...

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Theoretically, these requirements arise from the way you connect the measurement devices to the rest of the circuit. A voltmeter is connected in parallel, as you said. Say that you are trying to measure the voltage drop across a resistor $R$ through which passes a current $i$. If the internal resistance of the voltmeter is comparable to $R$, then the ...

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a resistor reduces the potential energy of the current across it, then the current that leaves the resistor will have less potential energy and thus, less pressure or voltage. The current does not have potential energy, the electric field does (the potential being exactly its potential as in the sense of differential forms). If one resistor only ...

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I know I'm a little late, but I'll take a shot at answering this for you. I'm actually very much a beginner at understanding electronics myself, so everyone: please keep me honest! There has been some criticism of your question, as it does not show a complete circuit. I need to agree with this, as any reliable calculations within a circuit require ...

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The current flows through both resistors. What works for me is to use the analogy of fluid through pipes under pressure. Imagine two huge tanks, connected by two small pipes of different sizes. The pressures in each of the two tanks are analogous to the two voltages. The pipes are analogous to the resistors. The fluid is analogous to the electrons. The ...

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Consider resistors $R_1$,$R_2$,$R_3$ connected in parallel (voltage $V$ is across them, $I$ is the total current from $V$. and $I_1$,$I_2$ and $I_3$ are currents in each resistor) The voltage across each resistor is same, which is $V$. So current would be same through it, even hundreds of resistances are connected in parallel. So if we add more resistances ...

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The solution is very simple. [EDIT: My diagram is almost similar to yours, just without the internal resistance r. I have talked about r in the last few lines. Anyway my diagram totally logical and valid for your problem.] See, your circuit sums up as follows: Now since the switch is open, you can consider that the lower branch of the circuit has been ...

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In the following situation you have the voltage source ensuring a potential difference V = 60 Volts between its terminals. The source's upper terminal is connected to the switch's upper terminal, so they have the same electric potential. The switch's lower terminal is connected to the resistor's upper terminal, so they also have the same electrical ...

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The problem here lies with the assumption that the resistance of the battery, wire and switch are negligible. There are two solutions that I can think of: 1) Because the two branches are connected in parallel, the branch with the resistor will still have $I_R=\frac RV$ current through it while the switch $S_1$ will have "infinite" current going through it ...

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To be clear, current is charge passing through a certain area per unit time. This does not imply a second parameter in the denominator of the formula for current dq/dt; just a guideline for how to measure dq. The larger the cross sectional area, the larger the perceived current will be. This is why resistance is seen to decrease for larger cross sectional ...

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Both holds true. If you use Ohm's law, you can easily see that $$i_1 R_1 = i_2 R_2$$. So, $$10 \times 1 = i_2 \times 0.2$$ gives $$i_2 = 50\,\mathrm{ampere}$$ Again by power conservation, $$V_{left} i_{left} = V_{right} i_{right}$$ And current in left loop will increase to be $i_{left} = 500\,\mathrm{ampere}$. As you can see, here both Ohm's Law as ...

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First of all, if you are dealing with a network of finite number of resistors, try redrawing it in some form in which you'll be able to recognize the parallel or series connections. Secondly, take a look at Delta-Y Transform which might be really helpful in some cases. If these fail, turn to Kirchoff's laws i.e. put a test generator between the points ...

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I'm assuming that you are reading the temperature of the resistor here. Now if $V$ is the voltage applied across a resistor $R$, the power generated by joule heating by the resistor would be $\frac{V^2}{R}$. But there also will be some heat losses through radiation and this will counteract the joule heating so that the resistor reaches an equilibrium ...

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The overall resistance of the circuit is 8 ohms. $$I = \frac{V}{R} \\ I = \frac{16}{8} \\ I = 2 \textrm{ A}$$ 2 A will flow through the first four ohm resistor, 1.33 A through the 12 ohm and 0.67 A through the 6 ohm resistor. You can now calculate the power dissipated in each resistor by using $P=I^2R$.

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QUESTION Current is constant throughout the circuit with a resistor hence we cannot say that the electron loses kinetic energy after passing through the load. SOLUTION Current throughout the circuit with a resistor is constant , no doubt about that. But to be fundamental, current in a circuit is set up by the electric field, not by electrons. For ...

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To maintain a current you need to "push" the charge through any obstacles on the path. If there is resistance against the current, then the "push" must be large enough to overcome this. The potential difference is this "push". Of course, as soon as the resistor has been passed, then a large "push" is no longer needed to make the current keep moving. Now ...

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Yes. there is a very simple method to determine that. If the two components are connected in such a way so that they have exactly one common terminal and no other component is connected to that common terminal then the two components are connected in series. The two components are said to be connected in parallel when they have two common terminals Any ...

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I don't know if this answers your question, but whenever I talk about in-series and in-parallel resistors I like the water analogy. In this analogy the battery is a pump that lifts the water from low (potential) to high (potential). The electric current is the water current, and the resistor is a wheel or a constriction in the pipe. So I have In-Series ...

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I played with this equation for a while, and it seems that $\exp(x^2)$ makes a better fit for these data; such relations are not unheard-of in theory, are they? Then I got: $$R\approx 4.044 \cdot \exp\Big[ ((T- 524.8)/200.8)^2\Big]$$ [Now I am sorry for the bashing that follows, but this is just to advertise my result.] This usually differs at most 2 ohms ...

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The equation, based on a least-squares analysis, is R = 2412exp[-0.01903T]. The data fit is very good through most of the range, but it is a bit off on the low temperature end. If necessary, this can be fixed with a weighted least-squares approach, but only at the expense of the curve fitting less well in other parts of the graph. If this is for a class ...

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