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The answer to each of the following questions is a single-digit integer ranging from 0 to 9. Darken the correct digit. A parallel plate capacitor is connected to a battery and when it is fully charged, the energy stored in the capacitor is U_(1). Now it is disconnected from the battery and then reconnected to the same battery but with the polarity reversed. During the time capacitor is again charged, heat H is found to get dissipated from the capacitor. Calculate H//U_(1). |
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Answer» `U_(1)=(1)/(2)CV^(2)` When the capacitor is connected to the battery with reversed polarity then additional charge that the battery needs to supply to the capacitor is 2CV in order to REVERSE the polarity of the capacitor, as CV is the magnitude of the charge on each plate. But note that due to REVERSAL of only the polarity, energy stored in the capacitor is not going to change. Hence, whatever work is done by the battery gets dissipated in the form of heat. `H=W_(b)=QV=(2CV)V=2CV^(2)` Hence, we can get `H//U_(1)=4` |
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