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An ac source is connected to a capacitor $C$. Due to decrease in its operating frequency:
A
capacitive reactance remains constant
B
capacitive reactance decreases.
C
displacement current increases.
D
displacement current decreases.
Explanation
Lower frequency raises $X_C$, so displacement current decreases.
Detailed Solution
Capacitive reactance $=\frac{1}{\omega C}=X_c$ (say)
On decreasing the operating frequency, $\omega$ reduces.
As $X_c$ is inversely proportional to $\omega$, the value of $X_c$ increases.
$\therefore I_C=I_D=\frac{V_o}{X_c}$
As $X_c$ increases, therefore displacement current $I_d$ decreases.