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note: I accidentally thought OP was asking about a series $LC$, not a series $LCR$. Including the $R$ changes the results here by making the infinities turn into large finite values. Suppose you hook your series $LC$ circuit up to a voltage source with frequency dependent phasor $\tilde{V}_s(\omega)$. Intuition First let's guess what happens. At low ...


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Say your series RLC circuit is excited by a constant current source producing current $I$ at angular frequency $\omega$. Since all 4 elements (R, L, C, and source) are in series, the current through any one of them is just the source current. Then for each element, $V_n=IZ_n$. For an inductor, the impedance Z is given by $Z=i\omega{}L$. For a capacitor, ...


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Please look here you can make a function of w for both voltage across inductor and capacitor, and you can check the neighbourhood of resonant freq.


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In a STM, you find the distance between the (atomically sharp) scanning tip and the surface at which tunneling starts to occur - that is, electrons "jump" the gap between the two and a current flows. How much current flows depends on the size of the gap, and the voltage between them. If you have no voltage, no current will flow regardless of the gap. If the ...


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In this case,if they are connected one at a time to the battery then, your reasoning is correct. But if both are connected at same time in series, then current will be same and electric field will also be same.


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Yet a voltmeter will provide a positive reading if you put the positive lead at the location with higher potential But this is precisely what one should expect given the quote in your question. Consider a conductor with resistance $R$, oriented vertically, and with a constant downward electric (conventional) current through. In this case, there is a ...


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Why is this? By convention. If you put the negative (black) lead at GND (or, e.g., battery minus) and you put the positive (red) lead at VDD (or, e.g., battery plus) the reading on the meter is positive. It's telling you how much higher in voltage the red lead is than the black lead. It's a convention.


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In response to your comment, there is a way to find the voltage without finding equivalent resistance. You can write the potentials at each junction point. Taking the points on the lower line to be zero potential, the potential gain on going up the last branch is $1$V. Then going left, add another $1$V. Then, current through the last branch is 1A and 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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There's an energy transfer whenever there is a change in potential, not potential difference. The (electric) potential, measured in volts, is the electric potential energy (EPE) of a unit charge at a particular point in the circuit. So, imagine a particle of unit charge travelling in the direction of current. It starts off with a higher potential (therefore ...


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But isn't this solution wrong? I think it should be wrong, because we are talking about AC here; That is, 220V is peak voltage, not mean voltage. I think that mean voltage is something below 220V, thus resistance is going to be below 484 Ohms. If you do not live in central or north america, 220V will be RMS, so your teacher is right.


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A few points of confirmation/correction: Yes the electrons flow in a direction opposite to the "conventional" current. No there is no "deficiency" if electrons - rather they have a different "potential" which is caused by the chemical reactions in the battery. No you don't have to invoke "surface charges" in the wire in order to understand current - ...


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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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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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First let's establish the situation in which the result actually holds. Voltage itself is only well defined in electrostatics, and this result only holds in a steady state. In an ideal battery, there is no energy loss inside the battery during operation, and in the steady state just as much charge flows into the battery as flows out of the battery, and ...


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Imagine a free-standing battery (not connected to any wires) and take a closed loop through the battery, out one terminal, and back in the other terminal. The total work done in moving a test charge around that loop must vanish. For this to happen, the change in electric potential outside of the battery must equal the negative of the EMF change within the ...


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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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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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