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If you add more resistors in series the effect will be the opposite of what you say: the battery will last longer. A battery has a certain rated capacity, written in mAh (milliamps times hours). Divide this capacity by the current you are drawing and you will get how much will that battery last in hours, at the same current draw. More resistance means less ...


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Schroedinger's Cat explains well why only part of the impendance is taken. Why are they considering a phase difference of $\phi$ ... Also, why are they taking modulus of Z Here's an algebraic explanation: For the RLC circuit we can write the total impedance in a general form, $$Z_T=R + jZ_r,$$ where $Z_r$ is the total reactive impedance of the $L$'s ...


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Why are they considering a phase difference of $\phi$? Your calculations are not totally correct. The voltages across different impedants $V_C,V_R,V_L$ have a phase relationship between them and hence the different impedances $Z_C,Z_R,Z_L$ are not directly linearly related as you have done. Consider the following phasor diagram: I hope it is clear ...


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Along the lines of what @CuriousOne said you can use Kirchhoff's laws to understand the current and voltage drop at each circuit element with any complexity you desire. Kirchhoff's laws deal with the fact of the total sum of currents and voltages in any circuit junction and loop, respectively, must equal zero. The second law essentially uses Ohm's law to ...


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I'm assuming that you are asking this question in context of an L-C circuit. The reasonant frequency of an L-C circuit is given by the formula $$f = \frac{1}{2\pi}\sqrt{\frac{L}{C}}$$ where L is the inductance of the inductor and C is the capacitance of the capacitor. Hence if any of these two values are changed the reasonant frequency of the circuit will ...


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Within a battery a chemical reaction is responsible for moving mobile electrons from one terminal to the other. The terminal from which the mobile electrons come from is called the positive terminal (deficit of mobile electrons) of the battery and the one that they go to is called the negative terminal (surplus of negative electrons). If there is no ...


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Suppose that you double the separation of the plates of a parallel plate capacitor of initial capacitance $C$ which is connected to a battery of emf $V$. The capacitance of the capacitor become $\frac C 2$ and the energy stored in the capacitor changes from $\frac 12 CV^2$ to $\frac 1 2 \frac C 2 V^2$. There is a decrease in the energy stored in the ...


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I had thought that for part A both light bulbs would begin glowing since the capacitor isn't charged, but i have no idea how to tell which one is brighter. I would reason about this in the following way. (1) This is a series circuit and so, the current through each circuit element (battery, bulbs, capacitor, switch) is identical. (2) The bulbs are ...


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The simple answer is that different people have different pain thresholds Your gender, your stress level, and your genes all contribute to your sensitivity to pain.


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All the above ranges of the given electric current is in A.C. Hence, even 1 milliamp of a.c. current is dangerous for us. The reason for this is that our body has a capacitive property which lets the a.c. current to pass through us and we fell a shock even at a very low current. But d.c. current of 1mA or even 1A don't appear to be dangerous. All this is ...


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You can solve this problem by using Kirchhoff's two laws. Kirchooff's current law tells you that since this is a series circuit the current in each part of the circuit will be the same all the time. So whatever happens to bulb $A$ will also happen to bulb $B$ as they both have the same current flowing through them. Using Kirchhoff's voltage law you have ...


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You're on the right track. You're probably familiar with how the current decreases exponentially after closing the switch. $$I(t)=I_0 e^{-t/\tau}$$ Where $\tau$ is the time constant of the circuit given by $\tau = RC$, and $R$ is the total resistance of the bulbs. So when the switch is closed, current will be a maximum, and the bulbs brightest. As time ...


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a. Immediately after the switch is closed, are either or both bulbs glowing? Explain. They will both glow as some current passes through them as the capacitor is charging. b. If both bulbs are glowing, which is brighter? Or are they equally bright? Explain. They are both equally bright, because an equal and opposite charge is flowing on to ...


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You have to keep this important thing in mind while dealing with problems of these kind. A capacitor behaves as a pure conductor immediately after connecting it with the circuit (i.e., at time t=0). As time passes the capacitor begins to lose its conductance exponentially and finally after a very long time (theoretically when time t tends to infinity) it ...


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You're not the first, nor the last, to find the phrase "power flow" somehow wrong. For example, from W J Beaty's article on electrical misconceptions: ELECTRIC POWER FLOWS FROM GENERATOR TO CONSUMER? Wrong. Electric power cannot be made to flow. Power is defined as "flow of energy." Saying that power "flows" is silly. It's as silly as saying that ...


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Well, if you think about electric power, which includes current (notion of flux), then you'll end with the conclusion that if there's no flow, there is no power.


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Well, you could use the mathematical definition of linear. Theorem 1 - A system is said to be linear if it satisfies the following principle of superposition: $$ \alpha H(x) + \beta H(y) = H(\alpha x + \beta y)$$ If $H$ is described as a linear operation (in this case the operation of resistances in a circuit) satisfy superposition. Now, if you take ...


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If you have all linear relationships, the outcome must be a linear one. If you have apples, do anything to them, you will still have apples


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If you have covered ac circuit theory you would have found out that for a series circuit the currents in each part of the circuit are in phase but the voltages across the components in the circuit may be out of phase with the current. For a resistor the voltage across it and the current through it are in phase whereas for a capacitor the voltage across it ...


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This question really boils down to the basics of AC circuits. If you check your theory, you will observe that the voltage (time-varying) applied to a purely capacitive (power source and capacitor) circuit lags the current by $90^0$. But the same voltage applied to a purely resistive (power source and resistor) circuit is in phase with the current. Hence, ...


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If the inner conducting shell is charged then the charges will reside on the outside of the inner shell as there can be no electric field inside a conducting shell. The charges on the outside of the inner conducting shell will produce a radial electric field and the outer conducting shell will find itself in that electric field. However the outer ...


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I want to charge a 12v 100ah battery which need 20 amp of current. You can charge the battery at any current proving it is not too high and it might be that the 20 amp is the maximum charging current? If the adapter gives a constant (regulated) 12 volts then you will not be able to charge a battery of the lead-acid type it will require more than 12 V ...


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For instance, why don't measure the ability to store something by the volume it takes so why not charge per unit volume. There is nothing wrong with you defining a parameter which is the "charge per unit volume" but after defining it then what are you going to do with it? So here you have a capacitor and its charge per unit volume is $3 \;\text{C ...


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I understand that capacitance is the ability of a body to store an electrical charge and the formula is $C = {Q \over V}$ Perhaps you just need to top thinking of capacitance as that. "Capacitance" sounds like "capacity", which leads to an intuitive trap like this: If I have a basket with a capacity of 2 apples, then a basket with more capacity can ...


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We Use $C=Q/V$ because those were useful things to measure. It's often easy to forget, but many of the equations we use are chosen because the work, and because other equations didn't work. Never underestimate that part of the reality. We don't use "charge per unit volume" because that number is not constant. You can charge a capacitor up without ...


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You can use a high vertical tube to store water in it (fill it from the bottom by pushing the water in) How much water can you store? It obviously depends on the pressure you apply to push it in. If you push harder, there will be more water stored. The tube is characterized not the amount of water, but by how easy it is to store the water. Its "capacity" ...


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A capacitor is used to store energy in form of electric fields. This electric field is created by charges on plates of capacitor. So, basically you are storing charge on capacitors. Let someone ask you how much charge you can store in your capacitor.What would you reply? Clearly , you reply " I may store 1mC or 100mC, depending on Potential difference ...


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Capacitance is "charge over voltage" – and one farad is "coulomb per volt" – because the capacity of capacitors (something that determines their "quality") is the ability to store a maximum charge on the plate ($+Q$ on one side, $-Q$ on the other side) given a fixed voltage. When you try to separate the charges, you unavoidably create electric fields ...


1

why would increasing voltage, while keeping charge constant, have any effect on the ability of a body to store charge. (1) Capacitors don't store charge, they store electrical energy. For a capacitor, it is understood that one plate has charge $Q$ while the other plate has charge $-Q$ so there is no net electric charge stored. (2) If you increase ...


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The definition of resistance of a component is $\text{resistance of component (R)} = \dfrac{\text{potential difference across component (V)}}{\text{current passing through component (I)}}$ This is not Ohm's law it is the definition of resistance. It so happens that for some components it is found that $V \propto I$ which is called Ohm's law. This is an ...


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I'm not sure exactly what you're asking for. But let's say there's an electric current flowing through a straight wire segment of length $l$, then the change in $\Delta\phi$, or $V$, would be defined by $$\Delta\phi = \int \mathbf{E}\cdot d\mathbf{l}$$ Because it is a straight wire, $$\Delta\phi = E\cdot l$$ But we have a definition of current $$I = ...


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Georg Ohm's original experiments, 1825, established that for a set temperature, the current through a specific length of a conductor was proportional to the potential difference applied. Ohm's law is empirical; it cannot be derived directly from Maxwell's equations as it depends upon material properties. It is violated by many materials, and even then ...


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a) Capacitance C increases b) Charge Q remains unchanged c) if charge Q is constant while C increases, that means voltage V decreases (C=Q/V). d) U decreases ( U=Q^2/2C)


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You are correct about inserting lightbulbs at X and Y. The same current flows through each branch, so the lightbulbs will be equally bright. However, the question is not whether there is a difference in the currents flowing through X and Y, but whether, if you connect a wire between them, any current will flow through it. Your objection, of course, will ...


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Batteries are made up of one or more electrochemical cells, arranged in series. In these cells an electrochemical Redox reaction takes place: $R+ze^{-} \to R^{z-}$ $O-ze^{-} \to O^{z+}$ Where $R$ and $O$ resp. are a reducing agent and oxidising agent. This transfer of electrons provides the EMF of the cell. As more current is drawn from the cell, the ...


2

For copper the temperature coefficient of resistivity is $3.9\times 10^{-3} \text{K}^{-1} $ and the temperature coefficient of thermal linear expansion is $1.6\times 10^{-4} \text{K}^{-1} $. They differ by a factor of about 24 so a change in temperature will cause a bigger change in resistance than in the linear dimensions of copper. Resistance is given by ...


0

Let $V_{X}$ be the voltage(potential) at X and $V_{J}$ the voltage(potential) at J. The potential at the positive terminal of E is $V_{X}$, the potential at the negative terminal of E is $V_{X}-E$. This is also the potential at the left terminal of r. Since the current through r and the galvanometer is zero, by Ohm's law, the voltage drop across r and the ...


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It seems to me that, it is not following Kirchhoff second law. In fact, if the voltage $V_{XJ}$ is equal to the emf $E$, Kirchhoff's voltage law (KVL) is satisfied only if the voltage across the internal resistance is zero: $$V_{XJ} = E - V_r,\qquad V_{XJ} = E \Leftrightarrow V_r = 0$$ But this is the case if the cell B current is zero $$V_r = r ...


0

In your layout, imagine the "antenna/you"-capacitor being parallel to the existing one, the antenna being the upper plate. Parallel capacitors add up their capacity. So how do you become the plate although you are not connected to the wires? The first step is to understand is that this setup (inductor + capacitor) will generate frequencies, as you could ...


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The player's hand acts as a grounded plate remembering that the player is a reasonable electrical conductor. The capacitor is part of an inductor-capacitor circuit, as you have shown above, which control the frequency of an oscillator. So what is missing is a clear indication that the bottom part of the circuit is connected to the earth/ground.


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The potentiometer is used to compare voltages by matching them. At the balancing point no current flows through a galavanometer between the potentiometer circuit and the test circuit, so it does not disturb the test circuit. The potentiometer is most sensitive when, at the balancing point, the slider or jockey is towards, but not too close to, the end of ...


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Rheostats are typically used in the electricity lab to "build" an adjustable generator from a constant-voltage generator, e.g. in order to measure the current-voltage characteristics of a component.


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Your confusion can be met with a practical scenario. You know that it'e the electrical energy that is carried from the battery to the load by the electrons. A connection wire has very good conductivity. A resistor obstructs the flow of current. Inside the resistor, the electrons lose their energy in the form of heat due to collisions with the atoms and only ...


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Resistors act to reduce current flow... I will give a counter-example to the claim that resistors "act to reduce current". Consider a $9V$ battery connected across a $100\Omega$ resistor; the battery current is $90mA$. Now, connect another $100\Omega$ resistor in parallel with the first. The battery current increases to $180mA$. Here's another ...


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[Due to a silly rule I could not format links, I would appreciate if someone with reputation edited those for me. ] Consider the following analogy to water pipes: Wires are like pipes already filled with water; the water resembles the movable charge, in case of metal electrons. Voltage is a pressure difference between two points, e.g. because one is higher ...


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What this means is that the resistor reduces the current compared to a circuit that didn't have the resistor in it. Say you have circuit with a bulb and a battery in which 0.5 A of current flows. If you then introduce a resistor in series with the bulb the current everywhere in the circuit will be less than 0.5 A. The current entering and leaving the ...


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Yes, but where is the problem? :) If you have a resistor in your series circuit, then the current is reduced - in the whole circuit, of course, not just in the resistor. The current is the amount of charges per time. There are just less, in a certain time, but they still have to pass through all the other elements of the circuit too.


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When you apply an AC voltage to a capacitor, current will flow into the capacitor, and back out again. As long as the absolute voltage on the AC generator is higher than on the capacitor, current will flow to increase the charge on the capacitor; and when it's smaller, current will flow to decrease the charge. Note that since there is an equal and opposite ...


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I believe that the potential at point X should be 12−60I and the potential at Y should be 12−30I. However, in the several problems that I've tried, this is not the case. Please explain, why? A proper way to write this is (in terms of the node voltages and branch currents) is $$V_X = 12V - I_X \cdot 60 \Omega$$ $$V_Y = 12V - I_Y \cdot 30 \Omega$$ ...


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If the wire has a non-zero resistivity, then there will be a finite resistance between the points $A$ and $B$, say $R_{AB}$. However, to determine the p.d. between these points you would have to know the resistance of the whole wire (or do you just mean that the wire between the points $A$ and $B$ has finite resistivity?), otherwise you can use Ohm's Law to ...



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