Question

A capacitor is connected across an ac source. suppose the capacitance of the capacitor is reduced by a factor of 2. what happens to the capacitive reactance of the capacitor?

Answer 1

capacitive reactance is given by


`\large X_(C)=(1)/(j\omega C) \rightarrow X_(C) \propto (1)/(C) \rightarrow \frac{X_{C_(2)}}{X_{C_(1)}} = (C_(1))/(C_(2)) \\~as ~the ~capacitence~ is ~reduced~ by~ 2~so ~C_(2)=(C_(1))/(2) \\ \frac{X_{C_(2)}}{X_{C_(1)}} =(2C_(1))/(C_(1))=2`
so capacitive reactance becomes double