# Tag Info

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This is not really a reasonable physical situation. The problem is that if the resistance truly is zero then the charge time is also zero and the current is infinite, which is not a reasonable conclusion. In physical cases, something must give: the resistance of the wires, or the internal resistance of the battery, will cease to be negligible before that ...

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the capacitance is infinite as capacitor can not store charge. the charge flows through the conducting wire. capacitor plates should be separated by insulating medium i.e. dielectric. capacitance of any form of capacitor is never zero.

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To be clear on the setup, we have an ideal battery (DC voltage source) and an ideal wire (zero resistance. In ideal circuit theory, asking what happens when one connects an ideal wire across an ideal voltage source is essentially asking what happens when $1 = 0$. However, if we allow that the wire has non-zero radius and non-zero length, then, when the ...

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It depends on inductivity: I suspect the circuit would explode, with or without capacitor. With just the battery - which also has zero inner resistance - the current would get infinite during the first 0 seconds. Except - while no energy is lost in heating wires - that would make the whole capacity of the battery.going into creating a magnetic field ...

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A few pointers (since this is homework): A capacitor of capacitance $C$ charged to a voltage $V$ has charge $Q$ such that $$Q=CV$$ The positive side of capacitor 1 has charge $+Q_1$, the negative side has $-Q_1$ etc. Then the total charge (on the two capacitors) after connecting the capacitors depends on the polarity of the connection you made. If you ...

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When a wire with no resistance is connected to the terminals of an ideal battery, will a current exist in the circuit? yes, it would be infinite If a capacitor is added to the circuit, will it be charged by the battery or will it remain uncharged? yes, the capacitor would be charged.

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The battery would be shorted in first case. In the second case: Although, resistance is never zero. You see that while charge on a charging capacitor varies with time as $$q=CV(1-e^{-\frac{t}{RC}})$$ When you set $R=0 \Omega$, the result is undefined. Although you can talk of $R\mapsto 0^{+}\Omega$ In that case, the capacitor will be charged instantly. ...

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I will give you hints how to do it. If you do it yourself, you will gain confidence. Charge will be conserved separately in the two plates you connect The charges will flow till potential difference across both capacitors is same. Note that equation $1$ will matter on which $2$ plates are connected. When you connect a resistor, the steady state will ...

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Your teacher is right. Although you missed a factor of $2$ in last line. The case you are saying is NOT equivalent to this one. How do you plan on filling dielectric $K_1$ on the lower half which is already filled with $K_2$? This makes no sense.

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It is essentially two capacitors in parallel. It will therefore have twice the capacitance than if everything else were held constant and one of the +Q plates were removed.

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http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-7-capacitance-and-field-energy/ Watch this video from 24:00 minutes till about 32:00 minutes (a total of about 8 minutes).

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Suppose first that $C_{5}$ is absent, that the voltage at the connection point of $C_{2}$ and $C_{3}$ is zero and that of the connection point of $% C_{1}$ and $C_{4}$ is $V(\omega )$. The impedance of a capacitor $C$ at the angular frequency $\omega$ is $$Z(C)=\frac{1}{i\omega C}.$$ Then \begin{eqnarray*} V(A) ...

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You need to assume how the current gets divided across the whole circuit. For example, take the current coming out of the main battery as $I$, then when it reaches the loop that you have selected, it breaks into $I_1$ and $I-I_1$. Do the same for all the loops, then apply Kirchoff's Rule. For capacitors, apply it exactly the same way you do it for batteries. ...

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In theory, provided that the surface area of the electrode remains constant and that the electrode material is conductive, the capacitance will not differ significantly when using different metals. This is just a consequence of the fact that for parallel plates, the capacitance only really depends on plate surface area. However, for modern electrochemical ...

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First of all that's $U$$=$$(1/2)$$C$$V^2$. But there is also another relation : $U$$=$$(1/2C)$$Q^2$. And I think your confusion will be gone if you realize when to use which relation. If you have a capacitor with a constant voltage source connected across it, then you use the first relation. But when a capacitor is already charged and is not part of a ...

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