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This is a good question and let us analyse a general situation and then come to your question. Let us consider the following diagram: Two conducting plates $A$ and $B$ are placed parallel to each other. Let us assume that we supply a charge $Q_1$ to the left plate and $Q_2$ to the right plate. The charge distribution in the surfaces $1$, $2$, $3$ and $4$ ...


5

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 ...


4

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 ...


2

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 $0$ ...


2

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 ...


2

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 ...


2

You have an electric circuit like this: (image from Electronics tutorials: LR series circuit) It is described by the following equations for voltages and current $$\begin{align} V_R(t)&=RI(t) \\ V_L(t)&=L\frac{dI(t)}{dt} \\ V_S&=V_R(t)+V_L(t) \end{align}$$ and the starting condition $$I(0)=0.$$ These equations can be solved and the solution is ...


2

At t=0, there’s no current and the voltage across the inductor is the battery voltage. The current in the inductor will rise linearly with time because inductor current is controlled by $V = L di/dt$. An inductor is not a resistor or wire, it’s it’s own kind of thing. Eventually the current will start to get big. A real battery has a maximum current it ...


2

Charges do no "accumulate" on a capacitor during charging. The battery does work to remove electrons from one plate and deposit them on the other. The plate from which the electrons are taken then has a net positive charge and the plate to which the electrons are deposited has a net negative charge. So charge is simply moved from one plate to the other. ...


1

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 $V_o$ ...


1

Electrons want to go to positive potential as this decreases their energy. So, the capacitor plate connected to the positive terminal of the battery will be positively charged. This is because the electrons in the capacitor plate will flow toward the positive battery terminal, leaving behind a positive charge on the plate. The same reasoning can be made ...


1

why does charge stored in capacitor remain constant. Because you disconnected the voltage source. It's meant to be implied that the capacitor is disconnected from all external circuits. Therefore there's nowhere for the charge to go. And since charge is a conserved quantity, that means the charge on the capacitor plate must remain constant. The surface ...


1

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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Is the electric field of a capacitor not always parallel and would I also have to account for edge effects or something similar? Yes, this is the key. The field lines of a capacitor are not perpendicular to the plate everywhere. They are approximately perpendicular far from the edges, but as you get closer to the edge that approximation progressively fails....


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