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I think that the problem here is that you haven't properly set up the circuit equations. So if we have a circuit with a capacitor and a bulb connected in series in an AC circuit Since the bulb and capacitor are series connected, the current through each is identical. Denote the series current phasor as $I_s$. Assuming the source is an AC voltage ...

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First think of initially putting charges $\pm Q$ uniformly on the two plates and then letting the system arrive at the final configuration in isolation. If fringe effects were ignored, the initial uniform charge distribution on the plates will remain unchanged. In this case, the potential difference between the two plates will be $V_\infty=Q/C_\infty$, where ...

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You are right. The second expression would work for the following diagram:

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A single Maxwell (for instance) BCAP0350 2.7v ultra capacitor that's about the size of a D cell has a capacity of 1300 Joules (1.3 x 10^3 J). It is extremely useful to use ultracaps to charge batteries if the nature of the power source is intermittent and high current (say, at 35 to 175 Amps, also within spec of the one I listed). Charge the ultracaps ...

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Electrical components are said to be in parallel if they are across the SAME potential. All the three capacitors are thus in parallel. Electrical components are said to be in series if the same current passes through them. If you decide to treat C2 and C3 together as one component, then C1 is in series with the combination of C2 and C3. but individually, ...

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When the current is turned on for 10 seconds the capacitors have gradually charged according to the equation Q(t)=CV(1-e^(-RC)) Now after 10 seconds when the switch is open the two capacitors act as two voltage sources in parallel and the current could be found out by superposition principle which is used when we have more than one source

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After the switch is opened, we have a series circuit - the current through each circuit element is identical and equal to $i(t)$. Choosing a clockwise reference direction for the series current, KVL clockwise yields $$\frac{1}{1\mu F}\int_{-\infty}^t i(\tau) d\tau + i(t) \cdot 1k\Omega + i(t)\cdot10k\Omega + \frac{1}{1000\mu F}\int_{-\infty}^t i(\tau) ... 0 There's a really awesome trick for problems like this. This is going to be a long post but the method presented makes problems like this really easy. The idea is to turn the series branch C_2, R_2 into an effective parallel R and C. See the diagram. The effective parallel values are denoted C_{2,p} and R_{2,p}. Parallel capacitances just add, so ... 1 Here the capacitors look like in series but they are not actually. Capacitors can be said to be in series only if they carry same amount of charge which is not the case here.If you calculate charges will come out to be different.Look at the time constants(R*C) for the first branch and second branch, they are 1ms and 10sec respectively. As switch is on ... 5 When the switch is opened, the circuit is the equivalent of this, so I think you can clearly see that the resistors are in series and so are the capacitors. It seems that you already understand how to calculate series resistance, so I'll show how to understand series capacitance. First, you probably already know that capacitance is defined as the ratio ... 0 The best way to go about these kind of problems, is to use Kirchoff's loops. Whenever there is ambiguity about parallel/serial, Kirchoff is the way, since you disregard the question of serial vs. parallel altogether. Also remember that while resistors observe:$$ \frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2} $$for parallel and,$$ R_T=R_1+R_2 $$for serial, ... 1 One expects the energy stored in the capacitor to transform like the zeroth component of the four-vector (U,\vec p). In its rest frame the field configuration around the capacitor has$$(U,\vec p)_\text{rest}=(U_0,\vec 0),$$and by the Lorentz transformation the moving observer will see$$(U,\vec p)_\text{moving}=(\gamma U_0, \gamma\vec\beta U_0),$$where ... 0 I'm not sure what your physics background is. The phenomenon you describe comes from the equation of continuity. This can be concisely mathematically expressed as$$ \oint \left( {\bf J} + \epsilon_0 \frac{\partial {\bf E}}{\partial t} \right) \cdot d{\bf S} = 0  In this equation ${\bf J}$ is an electrical current density (current per unit area) and ...

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Depends on the mechanical tension on the plates and the dielectric. If the force that accelerates the capacitor is applied at one of the plates, acceleration towards the second plate will push the plates closer together and increase capacitance. If the acceleration vector points the opposite way, it will decrease the capacitance. The effect is known to ...

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0 is a limit (!). But, kidding aside, gate dielectrics for modern CMOS devices are in the few nanometer thicknesses. Since these are (functionally) capacitors, d can certainly be small. But, high-k dielectrics had to be introduced for further scaling to reduce tunneling and thus leakage currents. The reason is that real materials have a dielectric ...

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1) You're asking about non ideal capacitors (see e.g. here), i.e. answer to your question is 'yes'. The formula assumes that capacitor's plates are the perfect conductors. 2) Oblique - will not work since you connect the plates, the last one can be decomposed into three capacitors joined in parallel.

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There is another reason for capacitors in electronics (but probably not fans) and that is Power Factor Correction

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Many fan motors are single phase, permanent split capacitor (PSC) induction motors. Single phase motors inherently have no starting torque and to get around this problem, engineers often "trick" the motor into thinking it is being supplied by 2 phases instead of one. This is done by adding a second set of windings to the motor that are physically offset to ...

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A capacitor is often used for "decoupling". The wires into any electrical appliance have inductance (because they are long and thin). This means that if there is a sudden increased demand in current, there will be a significant voltage drop. A capacitor can act as a "tiny battery" that briefly supplies this current while the main supply catches up. A fan ...

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Suppose we put charge on a conductor. The more charge we add to the conductor, the harder it gets to add more charge. That is, when we add charge, the potential at the conductor gets higher, and there will be a larger potential difference between the conductor and elsewhere; we will need a larger force to overcome that potential difference. We want to know ...

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