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 AC is doubled, we can follow these steps:

### Step 1: Understand the formula for impedance
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,
- \( X_L \) is the inductive reactance,
- \( X_C \) is the capacitive reactance.

### Step 2: Define the reactances
The inductive reactance \( X_L \) and capacitive reactance \( X_C \) are defined as:
\[ X_L = \omega L \quad \text{and} \quad X_C = \frac{1}{\omega C} \]
where \( \omega = 2\pi f \) is the angular frequency and \( f \) is the frequency of the AC source.

### Step 3: Analyze the effect of doubling the frequency
If the frequency is doubled, then:
\[ f' = 2f \quad \Rightarrow \quad \omega' = 2\omega \]
Now, substituting this into the expressions for \( X_L \) and \( X_C \):
\[ X_L' = \omega' L = 2\omega L = 2X_L \]
\[ X_C' = \frac{1}{\omega'} C = \frac{1}{2\omega} C = \frac{1}{2} X_C \]

### Step 4: Substitute the new reactances into the impedance formula
Now, substituting \( X_L' \) and \( X_C' \) back into the impedance formula:
\[ Z' = \sqrt{R^2 + (X_L' - X_C')^2} \]
Substituting the new values:
\[ Z' = \sqrt{R^2 + (2X_L - \frac{1}{2}X_C)^2} \]

### Step 5: Compare the new impedance with the original impedance
The original impedance was:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Now we need to analyze how \( Z' \) compares to \( Z \).

Since \( X_L \) is now doubled and \( X_C \) is halved, we can see that:
\[ Z' = \sqrt{R^2 + (2X_L - \frac{1}{2}X_C)^2} \]
This indicates that the term \( 2X_L - \frac{1}{2}X_C \) is greater than \( X_L - X_C \), leading to:
\[ Z' > Z \]
Thus, the impedance increases when the frequency is doubled.

### Conclusion
When the frequency of AC is doubled, the impedance of an LCR series circuit increases.

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