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1

$E=\frac{V^2}{r}t$ only holds if $I=\frac{V}{r}$, which only holds if the internal resistance $r$ is the only resistance in the circuit. Taking into account the load resistance $R$ we would get $I=\frac{V}{r+R}$, and hence $E=\frac{V^2}{r+R}t$ (assuming the battery's voltage is constant; otherwise it would become an integral). If you want to take the ...


2

Your analysis doesn't apply to what you are applying it to. $P=V^2/r$ is the power dissipated in the internal resistance of the battery. It is the power dissipated by the internal resistance if the battery is shorted. What you want is "the power supplied by the battery", which is the power dissipated in the external resistor: $P = V^2/R$, where care ...


2

A negative charge attracts a positive charge and repels an also negative charge. If you have many (negative) electrons in one end of a wire and non in the other end, then these electrons will be pushed away from each other towards the end from which they are not pushed away. This principle is the key - the electron-dense end is called the negative end, the ...


0

Your calculation is basically correct. If a battery is rated at 9.88WHrs it will produce a power of 1 Watt for 9.88 hours, or 8 Watts for 1.235 hours. However the power rating depends on what current the battery is producing. There will be an optimal current that gives the highest capacity, and the battery power rating is calculated for this optimum ...


2

No. Step number two already mentions electrons from the battery flowing into the negative terminal of the capacitor, giving the negative terminal a negative charge. Step number three is talking about electrons flowing out of the positive terminal of the capacitor, giving the positive terminal a net positive charge.



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