An ac voltage source that has a frequency fis connected across the terminals of a capacitor. Which one of the following statements correctly indicates the effect on the capacitive reactance when the frequency is increased to 4f? The capacitive reactance decreases by a factor of eight. The capacitive reactance increases by a factor of eight. The capacitive reactance is unchanged. The capacitive reactance decreases by a factor of four.
An ac voltage source that has a frequency fis connected across the terminals of a capacitor. Which one of the following statements correctly indicates the effect on the capacitive reactance when the frequency is increased to 4f? The capacitive reactance decreases by a factor of eight. The capacitive reactance increases by a factor of eight. The capacitive reactance is unchanged. The capacitive reactance decreases by a factor of four.
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![## Understanding the Effect of Frequency on Capacitive Reactance
An AC voltage source that has a frequency \( f \) is connected across the terminals of a capacitor. Which one of the following statements correctly indicates the effect on the capacitive reactance when the frequency is increased to \( 4f \)?
- \( \bigcirc \) The capacitive reactance decreases by a factor of eight.
- \( \bigcirc \) The capacitive reactance increases by a factor of eight.
- \( \bigcirc \) The capacitive reactance is unchanged.
- \( \bigcirc \) The capacitive reactance decreases by a factor of four.
- \( \bigcirc \) The capacitive reactance increases by a factor of four.
**Explanation:**
Capacitive reactance (\( X_C \)) is given by the formula:
\[ X_C = \frac{1}{2 \pi f C} \]
where:
- \( f \) is the frequency of the AC voltage source,
- \( C \) is the capacitance of the capacitor.
From this formula, we can see that capacitive reactance is inversely proportional to the frequency. Therefore, if the frequency increases, the capacitive reactance will decrease, and vice versa.
When the frequency is increased from \( f \) to \( 4f \):
\[ X_C \propto \frac{1}{f} \]
\[ \text{New } X_C = \frac{1}{2 \pi (4f) C} = \frac{1}{4} \left(\frac{1}{2 \pi f C}\right) \]
This indicates that the capacitive reactance decreases by a factor of four.
Hence, the correct statement is:
- \( \bigcirc \) **The capacitive reactance decreases by a factor of four.**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4eb8153a-3a83-4b3e-b817-a98859d3685f%2F2f776cee-d190-408a-804c-53b59f17a26d%2F0xlnncd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Understanding the Effect of Frequency on Capacitive Reactance
An AC voltage source that has a frequency \( f \) is connected across the terminals of a capacitor. Which one of the following statements correctly indicates the effect on the capacitive reactance when the frequency is increased to \( 4f \)?
- \( \bigcirc \) The capacitive reactance decreases by a factor of eight.
- \( \bigcirc \) The capacitive reactance increases by a factor of eight.
- \( \bigcirc \) The capacitive reactance is unchanged.
- \( \bigcirc \) The capacitive reactance decreases by a factor of four.
- \( \bigcirc \) The capacitive reactance increases by a factor of four.
**Explanation:**
Capacitive reactance (\( X_C \)) is given by the formula:
\[ X_C = \frac{1}{2 \pi f C} \]
where:
- \( f \) is the frequency of the AC voltage source,
- \( C \) is the capacitance of the capacitor.
From this formula, we can see that capacitive reactance is inversely proportional to the frequency. Therefore, if the frequency increases, the capacitive reactance will decrease, and vice versa.
When the frequency is increased from \( f \) to \( 4f \):
\[ X_C \propto \frac{1}{f} \]
\[ \text{New } X_C = \frac{1}{2 \pi (4f) C} = \frac{1}{4} \left(\frac{1}{2 \pi f C}\right) \]
This indicates that the capacitive reactance decreases by a factor of four.
Hence, the correct statement is:
- \( \bigcirc \) **The capacitive reactance decreases by a factor of four.**
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