When the frequency of AC is doubled, the impedance of an LCR series circuit

Options

A. is halved
B. is doubled
C. increases
D. decreases

Answer

To solve the problem of how the impedance of an LCR series circuit changes when the frequency of the AC source is doubled, we can follow these steps:

### Step 1: Understand the Impedance Formula
The impedance \( Z \) of an LCR series circuit is given by the formula:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
where:
- \( R \) is the resistance,
- \( X_L = \omega L \) is the inductive reactance,
- \( X_C = \frac{1}{\omega C} \) is the capacitive reactance,
- \( \omega = 2\pi f \) is the angular frequency.

### Step 2: Define Initial Impedance
Let’s denote the initial frequency as \( f \). The initial angular frequency is:
\[ \omega_1 = 2\pi f \]
The initial impedance \( Z_1 \) can be expressed as:
\[ Z_1 = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + \left(\omega_1 L - \frac{1}{\omega_1 C}\right)^2} \]

### Step 3: Define New Impedance After Doubling Frequency
When the frequency is doubled, the new frequency is \( 2f \), and the new angular frequency is:
\[ \omega_2 = 2\omega_1 = 4\pi f \]
Now, the new impedance \( Z_2 \) can be expressed as:
\[ Z_2 = \sqrt{R^2 + (X_L' - X_C')^2} \]
where:
- \( X_L' = \omega_2 L = 4\pi f L \)
- \( X_C' = \frac{1}{\omega_2 C} = \frac{1}{4\pi f C} \)

Thus, we can write:
\[ Z_2 = \sqrt{R^2 + \left(4\omega_1 L - \frac{1}{4\omega_1 C}\right)^2} \]

### Step 4: Compare the Two Impedances
Now we need to analyze how \( Z_2 \) compares to \( Z_1 \):
\[ Z_2 = \sqrt{R^2 + \left(4\omega_1 L - \frac{1}{4\omega_1 C}\right)^2} \]
We can see that \( X_L' \) increases (since it is multiplied by 4), while \( X_C' \) decreases (since it is divided by 4).

### Step 5: Determine the Effect on Impedance
To determine the overall effect on impedance:
- If \( X_L \) increases significantly more than \( X_C \) decreases, then \( Z_2 \) will be greater than \( Z_1 \).
- Conversely, if \( X_C \) decreases significantly more than \( X_L \) increases, then \( Z_2 \) could be less than \( Z_1 \).

However, generally, since the inductive reactance increases with frequency and the capacitive reactance decreases, we can conclude that:
\[ Z_2 > Z_1 \]

### Conclusion
When the frequency of the AC source is doubled, the impedance \( Z \) of the LCR series circuit increases.

---