# Tag Info

5

Wouldn't this inductor's emf counteract the discharging capacitor and actually charge it? / stop the capacitor from fully discharging? The inductor doesn't care about what the charge state of the capacitor is. All it cares about is how quickly the current through it is changing, and it generates a back-voltage according to the equation V=L*dI/dt. You ...

4

When a capacitor of capacitance C is charged to a voltage V, and discharged through a resistor R, then the current will decay exponentially: $$I = I_0 e^{-t/RC}$$ The voltage on the capacitor will follow the same exponential decay, $$V = V_0 e^{-t/RC}$$ To answer your question one would have to make some assumptions. You will have to do the calculation ...

4

By capacitor charge is meant the absolute value of the charge on each capacitor plate: $\mid Q \mid$. If the battery generates the potential difference $V$ and you connect the capacitor to the battery through a conducting wire, as shown in your picture, once the equilibrium is reached each plate of the capacitor will have a charge $Q = CV$, where $C$ is ...

4

The key to understand the issue is that between the upper and lower "corner" of the circuit the voltage is always zero, therefore no current will flow across $\mathrm C_5$ and $\mathrm C_6$. In "corners" I mean the points common to $\mathrm C_3$-$\mathrm C_4$ and $\mathrm C_1$-$\mathrm C_2$. These pairs of capacitors are effectively voltage dividers. ...

3

Just a simple answer - there's nothing to dissipate energy. If there were a resistor in the circuit, it would dissipate energy as heat. Inductors and capacitors don't dissipate energy. The energy just sloshes back and forth between being stored in the magnetic field, and being stored in the electric field. It's just like a spring-mass system, where energy ...

2

$E_1 = E_2$ . since $E$ is independent of dielectric as long as potential b/w plates is constant. $$E= = -\frac{dV}{dr}$$ So, it is independent of dielectric b/w it. So, correct statement would be $$E_1d = E_2d$$ $$Ed = Ed$$

2

The inductor never creates a current in the opposite direction. An inductor creates an EMF to counteract the changing B field(Lenz law). The B field is changing because the current in the inductor is changing. So effectively, the inductor resists changes in current. So initially, the capacitor tries to discharge strongly but is slowed down by the ...

1

As you may know, it takes infinite time to charge a capacitor. So, the time when the capacitor is 100% charged never comes. Thus, we require a Time Constant to help us understand the time when the capacitor has got a decent amount of charge and after which the rate of charging becomes really slow and thus charging further is not of much use. You may also ...

1

You seems to assume both capacitors has the same plate separation $d$. So, lets assume that. Assume there is no dielectric material. Therefore, nicely $Ed = Ed$ in both capacitors. Which is nice. :). Now, I think I understand your confusion. Have an isolated capacitor with electric field inside plates of $E$. Insert dielectric $K$. Under this case, the ...

1

The difference is that batteries chemically "pump" electrons from one side to the other. There is a small amount of charge separation in a battery even when it is not connected to a circuit. This charge creates an electric field that opposed the chemical action of the battery to prevent further charge separation. This makes the battery act somewhat like a ...

1

Perhaps this is what you are looking for: Screen capture: http://www.falstad.com/circuit/ The default circuit, as shown, is an LRC circuit. On the Schematic: Gray is 0V Green is Positive Voltage Red is Negative Voltage The yellow dots are a visualization of current: positive holes. The graphs along the bottom, from left to right, are for the ...

1

I asked a somewhat different, yet similar question.Hope this helps! Why is an $LC$ oscillator lossless, but $C V^2 / 2$ energy is lost to a capacitor connected to an ideal voltage source?

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