AC signal, then reactance of that circuit is purely capacitive. The capacitive reactance is \[X=\frac{1}{\omega C}=\frac{1}{2\,\pi \,f\,C}\] \[(\omega =2\,\pi \,f)\] or \[X\,\,\propto \,\,\frac{1}{f\,C}\] \[\THEREFORE \] \[\frac{X'}{X}=\frac{fC}{f'C'}=\frac{f\times C}{2f\times 2C}\] or \[\frac{X'}{X}=\frac{1}{4}\] or \[X'=\frac{X}{4}\]

"> AC signal, then reactance of that circuit is purely capacitive. The capacitive reactance is \[X=\frac{1}{\omega C}=\frac{1}{2\,\pi \,f\,C}\] \[(\omega =2\,\pi \,f)\] or \[X\,\,\propto \,\,\frac{1}{f\,C}\] \[\THEREFORE \] \[\frac{X'}{X}=\frac{fC}{f'C'}=\frac{f\times C}{2f\times 2C}\] or \[\frac{X'}{X}=\frac{1}{4}\] or \[X'=\frac{X}{4}\]

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The reactance of a capacitor of capacitance C is X. If both the frequency and capacitance be doubled, then new reactance will be: [AIPMT 2001]

NEET Physics in NEET 1 year ago

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[d] If a capacitor of capacitance C is connected with an AC signal, then reactance of that circuit is purely capacitive. The capacitive reactance is \[X=\frac{1}{\omega C}=\frac{1}{2\,\pi \,f\,C}\] \[(\omega =2\,\pi \,f)\] or \[X\,\,\propto \,\,\frac{1}{f\,C}\] \[\THEREFORE \] \[\frac{X'}{X}=\frac{fC}{f'C'}=\frac{f\times C}{2f\times 2C}\] or \[\frac{X'}{X}=\frac{1}{4}\] or \[X'=\frac{X}{4}\]

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