Question

If a variable frequency ac source is connected to a capacitor then with decrease in frequency the displacement current will

• A. increase
• B. decrease
• C. remains constant
• D. first decrease then increase

Answer 1

Answer: (B)

decrease

Answer 2

Current through capacitor, $I = \frac{E}{X_{C}} = \frac{E}{\frac{I}{\omega C}}$ $= \omega CE = 2\pi\upsilon\,CE \quad$ or $\quad I \propto \upsilon$. $\therefore\quad$ decrease in frequency $\upsilon$ of ac source decreases the conduction current. As displacement current is equal to conduction current, decrease in $\upsilon$ decreases displacement current in circuit.