# Tag Info

## Hot answers tagged electric-circuits

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Batteries contain various liquids that are important for the voltage to be produced. Sometimes, the liquid – even water – may turn to gas and it is permanently lost, along with the capacity. Sometimes, the liquid just moves to one side of the battery which is also bad. Shaking a weak battery may homogenize the concentration of the liquid across the battery. ...

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The missing piece here is that the temperature of the resistor is a function of the current. Your equation should perhaps read $V = I\,R(T(I))$. Does that help?

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There is a commonly used analogy for electric circuits called the hydraulic analogy. This imagines the electrons as water and the wires as pipes. The voltage is equivalent to the water pressure and the current is equivalent to the water flow rate. Start with a DC current and imagine the water is doing work by flowing through a water wheel: This is all ...

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The cathode ray tube has had the air pumped out. Electrons scatter off oxygen and nitrogen molecules so if you fired an electron beam in air it would be scattered in a short distance. The distance would depend on the beam energy, but it's a lot shorter than 100m. The range of electrons from beta radiation in air is around a metre. You could argue that ...

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Two capacitors in parallel have the same voltage. Two capacitors in series have the same charge. Simplify the problem to two capacitors in series (each started life as two capacitors in parallel) - what is the ratio of their voltages. Then use $Q=CV$ to figure the charge on each pair; finally distribute the charge on the elements of each pair according to ...

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Hint in a parallel circuit, the voltage across each resistor is the same. What is the voltage across the $4 \Omega$ resistor?

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Reluctance = $\dfrac{l_e}{\mu A_e}$ where..... mu is the absolute permeability of the material, $\mu_0 \mu_r$ $l_e$ is the circumference of a circle at a radius r and $A_e$ is a small cross sectional area. The circle I refer to only relates to the cross section of the torus and r is the radius from the centre (where the wire is). All these reluctances ...

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An easy way to prove Ohm's law for electric fields that aren't constant is to first assume that the electric field is approximately constant over short lengths, just like $E=dV/dL$ suggests. Using that, you can derive Ohm's law for short lengths of material, $dV=IdR$. We'll assume that "current in = current out", which is true at steady-state. This allows us ...

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Assuming for now that this is homework, I'll provide this hint: the voltage on the 8.73 $\mu$C capacitor is not 21.9 V. Don't forget that that voltage has to be distributed among all of the components.

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In principle, it is possible, using, e.g., high-current relativistic electron beams - please see, e.g., the review http://arxiv.org/abs/physics/0409157 . @John Rennie offers reasonable arguments, but the very real problems he mentions can be overcome - I don't have time to describe the specific mechanisms (see the review). In experiments, propagation length ...

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To start with one could have an ac current never grounded anywhere , for a household generator for example. The reason one grounds at the generator is for safety so the ground can pick up any miss chance, as it is a practically infinite sink for electrons. Only one of the two lines can be grounded of course :). It was found though that due to capacitences ...

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From a circuit theory perspective, an inductor is effectively a short-circuit (wire) at DC while a capacitor is an open-circuit. Thus, any parasitic capacitance must be in parallel since, if it were in series, an inductor would be an open-circuit at DC. From an AC perspective, if the parasitic capacitance were in series, the inductor would appear ...

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You have a function: $V(T, i) = i \cdot R(T)$ and you should get $\dfrac{dV}{di}$. $T$ doesn't change when you vary $i$ and $R(T)$ doesn't too, so it can be considered as a constant comparing to variable $i$. Fix $T$ at some generic value, for example $a$, doing this you get $R(T = a) = R_a$ So your function is reduced to $V(i) = i \cdot R_a$. Now you ...

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I think the shaking action causes the batteries to move slightly, thus 'scraping' the contact area on the batteries and the contact elements in the remote. This improves the conductivity, and so on. Certainly I've had success opening the battery compartment and physically rotating the batteries and/or scraping the contacts gently. I think it's unlikely ...

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