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While one can specify a voltage at a point, it must always be with respect to another point, sometimes referred to as common, ground or reference point where the voltage is arbitrarily assigned a value of zero. The zero voltage point is typically chosen at the negative terminal of the voltage source. For the simple circuit below, I have labeled points A, B,...


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Using imaginary numbers for current in reactive components just happens to make the maths a lot simpler. In AC circuits there is typically some phase difference between the voltage and the current. Manipulating these quantities without the use of complex numbers, but instead just keeping track of the phase differences (such as the power factor), is a right ...


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An RLC circuit satisfies$$L\ddot{I}+R\dot{I}+C^{-1}I=\dot{V}.$$To solve this with AC voltage such as $V=V_0\cos\omega t,\,V_0\in\Bbb R$, it's convenient to take the real part of a complex choice of $I$ for the case $V=V_0\exp j\omega t$. Substituting $I=I_0\exp j\omega t,\,I_0\in\Bbb C$ gives$$I_0=\frac{j\omega V_0}{C^{-1}-\omega^2L+j\omega R}.$$The special ...


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in Ohm's law we are measuring voltage at a point so work done should be very small since there is almost no distance This is the key mistake. In Ohm’s law the voltage is the voltage across the resistor. In other words, the voltage in Ohm’s law is measured at two points. One point is always the point where the current is entering the resistor and the other ...


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Expanding on Pieter's answer, consider a transmission line (coaxial cable) lumped model that includes resistance, capacitance, inductance, and conductance. You can easily google this model to find detailed descriptions. However, below I show a simple diagram of the lumped model: The salient feature of the model is that resistance (R) and inductance (L) ...


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Short Version: When the resistance in the above circuit is zero, the system will never attain equilibrium. Long Version: Any time you want to understand what happens with a zero resistance you need to start with a small but non-zero resistance and see what happens in the limit of $R \to 0$. by John Rennie sir in chat. When a capacitor of ...


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Electrons inside solids have a very random motion. They keep on bumping around here and there. What happens when you put a battery? The motion of these electrons is only slightly disturbed. Free electrons inside the metal move at large speeds, a battery sets up an electric field inside the solid which tries to push the one side but the acceleration caused is ...


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$R$ is not in series with the 50 Ohm resistor. It is in series with the parallel combination of the 50 Ohm resistor and the resistance of the lamp. Hope this helps


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When the switch is closed, we can easily see that the sum of voltage drops across the right hand side loop must be $0$ by Kirchoff's loop rule. So we have $$V_{R_2}+V_C=0$$ $$-R_2i-\frac qC=0$$ $$\dot q=-\frac1{R_2C}q$$ So you can see that the time constant only depends on $R_2$. As mentioned in the comments, you can also understand this by noting a $0$ ...


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The electrical power does not only come from a flow of electrons, but also because something pulls them : an electric field. We model this by defining the electric power received by a component as being the product of the current flowing through it by the voltage applied to it (in the opposite direction, by convention). As for whence the energy came from ...


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A capacitor (with capacitance $C$) is fully described by the differential equation between current $I(t)$ and voltage $V(t)$: $$I(t)=C\frac{dV(t)}{dt} \tag{1}$$ Suppose you have an AC voltage with frequency $\omega$ connected to the capacitor. By using the complex calculus this is $$V(t)=V_0 e^{j\omega t} \tag{2}$$ Then, by plugging voltage (2) into ...


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Sometimes notation may induce misunderstandings. It is true that in many formulas voltage appears as $\Delta V$, making clear that it refers to the difference of voltage between two points. However, arrived at Ohnm's law, it is usual to find it expressed as $$ I=\frac{V}{R},$$ thus inducing to think that only voltage at one point is present. Actually in ...


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The current in the rightmost 5 Ohm resistor has no influence on the current in the 10 Ohm resistor and vice versa because the voltage across the 5 Ohm resistor is fixed at 10 volts. It can therefore be ignored.


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definition of voltage as work done moving unit of electric charge between two points in electric field That's almost okay, but what's missing is the concept of electrical potential. And technically, you should say "work per unit charge" to get the units correct (volt = joule per coulomb). That's often not discussed in a lower level (high school) first ...


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First, this has nothing to do with the capacitor. Regardless of what load you connected there (capacitor, inductor, some complicated transistor circuit, or whatever), you'd find the Thevenin equivalent of the source the same way. Second, when you ask for the Thevenin equivalent resistance of the circuit, you're asking what's $\frac{dV_o}{dI_o}$, where $V_o$ ...


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Here's how I understand what you are saying. Correct me if I'm wrong. A circuit might be rated for 25 A, but the appliance uses only 2 A. That leaves 23 A left over. You wonder why those left over amps don't heat things up. A resistor, an appliance, or anything else, heats up only when current is flowing through it. Massive amps do not come from the ...


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In all aspects of nature, movement (and more general stuff too) arises due to what physicists call potential differences. That's why one calls them potential: Because they have de potential of creating kinetic energy. In the case of gravitational potential energy, for instance, it can be converted to kinetic energy by getting rid of normal reaction forces (i....


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Here's one way to think about it. Consider the case that one cell is connected across a resistor. Stipulate that the cell has a capacity of 1 amp-hour and that the cell delivers 1 amp of current to the resistor when first connected. You should expect that the cell will discharge in roughly 1 hour. Now, disconnect the discharged cell, and connect to the ...


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Because these resistance pathways are in parallel (perpendicular to the axis of the cylinder).


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First, you have to look at your starting assumptions. It is not generally true that there is more power in a circuit with a smaller resistance. So we need to examine the circumstances where it is true and where it is not true to see if we can discover any hints. It is not true that there is more power in a circuit with a smaller resistance when the two ...


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Except for gradual energy losses that occur inside all batteries (giving them a limited shelf life) there is no energy lost by your 9 volt battery unless it does work to deliver current (charge) to a circuit connected to its terminals. The potential difference, or voltage $V$ between two points is the work (energy) required per unit charge to move the ...


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When charges move due to an electric field (i.e. down the potential gradient) the electric potential energy of the system decreases. When you put in the second capacitor there is a brief moment where an electric field is present that causes excess charges to move to the other plates. Hence you have less energy stored. Or you can view it as doubling the ...


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Yes, even assuming a perfectly conductive circuit, there would be some inductance (you may not have studied this yet) which would create an oscillating LC circuit, which would radiate. Some early spark gap radio transmitters worked like this.


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There is a lot of random thermal motion of charges, but on average their speed (drift velocity) through a resistor is constant. Around a given series circuit the current is constant with respect to position, so the average speed depends only on the product of the cross sectional area and the free electron density. Since that is constant across a resistor the ...


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