Question

Same current is flowing in two different a.c. circuits. First circuit contains only inductance and second contains only capacitance. If the frequency of a.c. is increased in both circuits, the current will

• A. increase in the first circuit and decrease in second.
• B. increase in both circuits.
• C. decrease in both circuits.
• D. decrease in first circuit and increase in second.

Hint:

Inductors block high frequencies (low-pass), while capacitors pass high frequencies (high-pass).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks about the change in current when frequency increases, assuming the applied voltage remains constant. Current $I$ is inversely proportional to reactance $X$.
Step 2: Key Formula or Approach:
1. Inductive reactance $X_L = \omega L = 2\pi f L$.
2. Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$.
3. Current $I = \frac{V}{X}$.
Step 3: Detailed Explanation:
Circuit 1 (Pure Inductor):
As frequency $f$ increases, $X_L \propto f$ increases.
Since $I_1 = \frac{V}{X_L}$, an increase in $X_L$ leads to a decrease in current.
Circuit 2 (Pure Capacitor):
As frequency $f$ increases, $X_C \propto \frac{1}{f}$ decreases.
Since $I_2 = \frac{V}{X_C}$, a decrease in $X_C$ leads to an increase in current.
Step 4: Final Answer:
The current will decrease in the first circuit and increase in the second circuit.