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If biased around DC, the capacitor stores energy as the voltage increases in magnitude (either positive or negative), and the capacitor gives up energy as the voltage returns to zero. Thus 1/4 cycle is store, 1/4 cycle is discharge, this occurring for each 1/2 cycle. If biased asymmetrically, then the timing changes.


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Yes! It's true or false. When connected across a sine-wave AC supply an ideal capacitor stores energy while the voltage is increasing in magnitude and releases energy when the voltage is decreasing in magnitude. so half the tlong period will ne equal ime it's storing and half the time releasing. But the it's not a continuous half of the AC cycle, it's ...


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No energy is lost in a purely reactive load, which a capacitor is. Over time for any repeating voltage signal with no average bias, the cap’s average charge will be zero. That is, the net charge is the integral of applied voltage over time. Zero integral => no charge. Now, whether it ‘returns’ its energy in 1/2 cycle depends on the waveform. If it’s ...


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True. The energy stored in a capacitor is given by \$ U = \frac {1}{2}C V^2 \$. If the average energy is increasing then V would be increasing too. Figure 1. An example circuit. Physics and radio electronics. The circuit of Figure 1 gives a common example (which doesn't quite match the charge for ½ cycle requirement of the title as the charge ...


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yes it is true.because the basic working of capacitor is that it charges for one half cycle and discharges the energy that stored in 1 half cycle.for ideal capacitor net energy is Zero.


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FALSE! But it's a trick question. (Or, perhaps the author has misconceptions about how capacitors work, and their book has errors?) Was the answer supposed to be false? Yet the net energy is actually zero, that part's true. The question's reasoning is wrong, the number "½" is wrong, so the answer given in the back of the book had better be "false."


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