Question

$\text{Assertion}$:
When a variable frequency $AC$ source is connected to a capacitor, displacement current increases with increases in frequency.

$\text{Reason}$:
As frequency increases, conduction current also increases.

• A. Both $\text{Assertion}$ and $\text{Reason}$ are correct, and $\text{Reason}$ is the correct explanation for $\text{Assertion}$
• B. Both $\text{Assertion}$ and $\text{Reason}$ are correct, but $\text{Reason}$ is not the correct explanation for $\text{Assertion}$
• C. $\text{Assertion}$ is correct, but $\text{Reason}$ is incorrect
• D. Both $\text{Assertion}$ and $\text{Reason}$ are incorrect

Answer 1

The correct option is A Both $\text{Assertion}$ and $\text{Reason}$ are correct, and $\text{Reason}$ is the correct explanation for $\text{Assertion}$

The impedance $Z$ of a pure capacitive circuit is given by

$Z=X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

As the frequency increases, the impedance of the circuit decreases.

Now, the conduction current flowing in the circuit is given by

$I_C=\frac{V_S}{X_C}$

$\Rightarrow I_C\propto \frac{1}{X_C}\propto f$

So, as $X_C$ decreases conduction current $I_C$ increases.

And we know that the value of displacement current is equal to the value of conduction current, therefore, the displacement current will also increase with increase in frequency.

Hence, $\text{Assertion}$ is correct, $\text{Reason}$ is correct and $\text{Reason}$ is the correct explanation for $\text{Assertion}$.

Therefore, option $\left(\text{A}\right)$ is correct.