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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?

Respuesta :

naveen1997
capacitive reactance is given by 

[tex]\large X_{C}=\dfrac{1}{j\omega C} \rightarrow X_{C} \propto \dfrac{1}{C} \rightarrow \dfrac{X_{C_{2}}}{X_{C_{1}}} = \dfrac{C_{1}}{C_{2}} \\~as ~the ~capacitence~ is ~reduced~ by~ 2~so ~C_{2}=\dfrac{C_{1}}{2} \\ \dfrac{X_{C_{2}}}{X_{C_{1}}} =\dfrac{2C_{1}}{C_{1}}=2 [/tex]
so capacitive reactance becomes double

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