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A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance is reduced.

Answer

When \(\mathrm{DC}\)  voltage is applied than the capacitive reactance that is the resistance offered by the capacitor will be \(\frac 1{2\mathit{\pi fC}}\) where \(f\) is the frequency and \(C\) is the capacitance.
For \(\mathrm{DC}\)
\(f=0\)
So capacitive reactance \(=\) infinite
Therefore after charging the capacitor will not allow the DC current to flow, it will block DC. As lamp is connected in series so it will not shine and there will be no change in brightness by reducing capacitance.
But in case of AC, by decreasing capacitance the capacitive reactance increases So
\(I=\frac{\mathrm{Voltage}}{\mathrm{Reactance}}\)
So current decreases and the bulb will shine less brighter.
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