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At equilibrium, the field inside an ideal conductor is zero. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gausur.html#c2 A charge moving through such a conductor neither gains nor loses energy. We can't attach an ideal conductor to an ideal voltage source. Something has to give. There will be a voltage drop along a real wire due to non-zero ...


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Let me first take a little detour away from this circuit to particle accelerators. If you have some electrons in vacuum and a potential set up between two points (exactly the same as saying you have an electric field set up) you can accelerate your electrons. If you move a single electron through $1V$ of potential the electron gains $1eV$ of energy where ...


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You should not simply add the errors, you should sum them squared in case of $y=AB/C$. $dy/y=\sqrt{ (dA/A)^2 + (dB/B)^2 + (dC/C)^2 }$ This comes from the partial derivation of the function $y$ by all the components and weighting them by the uncertainty: $dy^2=(dA\frac{\partial y}{ \partial A})^2 + ... $ The square is there because you treat the different ...


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I think the key thing missing in your thinking is that the energy drop across a resistor is not just determined by the properties of the resistor, but also by how much current flows. The cool thing is that no matter what resistors you put in, the current that flows is such that the potential will fall all the way back down. The reason for this is that ...


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This is a common question. The issue is that the "Q" in $i = dQ/dt$ is not the same as the $Q$ that represents the charge on the capacitor. The variable $Q$ in use here is simply the charge on the capacitor. No problem. When the capacitor discharges the quantity of charge that is introduced into the circuit after a time $\delta t$ has elapsed is ...


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I recently answered a similar question here. The ideal capacitor equation $$i_C = C\frac{dv_C}{dt}$$ assumes the passive sign convention which means that the reference direction for $i_C$ is into the positive labelled terminal. When you write $$iR = v_C$$ it is necessarily the case that $$i_C = - i$$ To see this, assume that both positive labelled ...


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A good reference was given in an answer to a related question: Cserti 2000 (arXiv preprint, whose numbers I'll be referring to) solved a number of generalizations of the 2D lattice problem. For a $d$-dimensional lattice, the resistance between the origin and the point $(l_1, \ldots, l_d)$ is given by eq. 18 in that paper: $$ R(l_1, \ldots, l_d) = R_0 ...


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For your circuit, $V = I\cdot R$. You are plotting (unusually) R along the X axis and $\frac{1}{I}$ along the Y axis, so the slope is $\frac{1}{V}$. Now the fact that this slope is a straight line tells you that the voltage is constant. This means that (over the range of your experiment) your voltage source has a low internal resistance. Imagine for a ...


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Conductance is the extrinsic property while conductivity is the intrinsic property. This means that conductance is the property of an object dependent of its amount/mass or physical shape and size, while conductivity is the inherent property of the material that makes up the object. No matter how the object changes in terms of shape/size/mass, as long as it ...


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If it's a simple circuit where Ohm's law applies, then we should get $$V=IR$$ so we see that $$V/I = R$$ $$1/I=R/V$$ $$1/I = (1/V) \times R$$ The gradient should then be $1/V$. Seems like a slightly bizarre plot but if you got a straight line then that makes the maths simple at least!


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The bleeder resistor is across the capacitor so they have identical voltage across.


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I think the approach given by jaromax is correct (+1, I also get 1.4$\Omega$), whereas the formula quoted in the linked question should not be used if the measurements of $R_1$ and $R_2$ are independent and (slightly) overestimates the total uncertainty. However, I am adding this answer because the approach you adopted based on percentage errors is ...


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So what exactly happens to the potential inside the resistor ? Unlike the ideal conductors, for which an electric field cannot exist inside, there is an electric field through the resistor body when there is a current through. And, as you may know, the rate of change in electric potential is related to the value of the electric field. Thus, the ...


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Trying to address this misconception: I start of with a resistance of 1 ohm by the wire and 6 amperes which result in 6 volts. When I meet the resistor however the resistance increases to let's say 3. Does the current decrease at the same rate the resistance increases? So if the resistance goes down to 3 will the current be 2 so that in the end I have a ...



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