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

Explanation

To solve the question of how the displacement current changes with a decrease in frequency when a variable frequency AC source is connected to a capacitor, we can follow these steps:

### Step 1: Understand the Concept of Displacement Current
Displacement current is a concept introduced by James Clerk Maxwell to account for changing electric fields in situations where there is no actual current flow, such as in a capacitor. The displacement current \( I_D \) can be expressed in terms of the voltage across the capacitor and the capacitive reactance.

### Step 2: Write the Formula for Displacement Current
The displacement current \( I_D \) can be related to the root mean square (RMS) current \( I_{RMS} \) through the following equation:
\[ I_{RMS} = I_D = \frac{V_{RMS}}{X_c} \]
where \( V_{RMS} \) is the RMS voltage across the capacitor and \( X_c \) is the capacitive reactance.

### Step 3: Express Capacitive Reactance
The capacitive reactance \( X_c \) is given by:
\[ X_c = \frac{1}{\omega C} \]
where \( \omega = 2\pi f \) (angular frequency) and \( C \) is the capacitance of the capacitor.

### Step 4: Analyze the Effect of Frequency on Reactance
As the frequency \( f \) decreases, the angular frequency \( \omega \) also decreases. Since \( X_c \) is inversely proportional to \( \omega \), we can conclude that:
- If \( \omega \) decreases, then \( X_c \) increases.

### Step 5: Relate Changes in Reactance to Displacement Current
Since \( I_D \) is inversely proportional to \( X_c \), if \( X_c \) increases, then \( I_D \) must decrease. Therefore, we can summarize:
- As the frequency decreases, the displacement current \( I_D \) decreases.

### Final Conclusion
Thus, the displacement current will decrease with a decrease in frequency.

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