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(B) The capacitive reactance $X_{c}$ is given by $X_{c} = \frac{1}{2 \pi f C}$.The conduction current $I_{c}$ in the circuit is given by $I_{c} = \fra

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It will decrease.

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(B) The capacitive reactance $X_{c}$ is given by $X_{c} = \frac{1}{2 \pi f C}$.The conduction current $I_{c}$ in the circuit is given by $I_{c} = \frac{V}{X_{c}} = V(2 \pi f C)$.According to Maxwell's modification of Ampere's law,the displacement current $I_{d}$ is equal to the conduction current $I_{c}$ in the circuit,so $I_{d} = I_{c} = 2 \pi f C V$.Since $I_{d} \propto f$,if the frequency $f$ decreases,the displacement current $I_{d}$ will also decrease.

$A$ variable frequency $AC$ source is connected to a capacitor. How will the displacement current change with a decrease in frequency?

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