Question

If a variable frequency ac source is connected to a capacitor then with decreased 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,

$I = \frac{E}{X_{C}} = \frac{E}{\frac{1}{\omega C}} = \omega CE = 2\pi\upsilon CE$ or $I \propto \upsilon.$

$\therefore$ decrease in frequency $\upsilon$ofj ac source decreases the conduction current. As displacement current is equal to conductiong current, decrease in $\upsilon$ decreases displacement current in circuit.