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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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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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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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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 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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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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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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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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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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