To address the question regarding the effect on the capacitive reactance when the frequency is increased to 4 times its original value, let's recall the formula for capacitive reactance ([tex]\(X_c\)[/tex]):
[tex]\[ X_c = \frac{1}{2 \pi f C} \][/tex]
where:
- [tex]\(X_c\)[/tex] is the capacitive reactance,
- [tex]\(f\)[/tex] is the frequency of the AC voltage source, and
- [tex]\(C\)[/tex] is the capacitance of the capacitor.
Given the initial frequency [tex]\(f\)[/tex], the capacitive reactance [tex]\(X_c\)[/tex] can be expressed as:
[tex]\[ X_c = \frac{1}{2 \pi f C} \][/tex]
Now, let's increase the frequency to [tex]\(4f\)[/tex]. The new capacitive reactance ([tex]\(X_c'\)[/tex]) would then be:
[tex]\[ X_c' = \frac{1}{2 \pi (4f) C} = \frac{1}{8 \pi f C} \][/tex]
To find the relationship between the new capacitive reactance [tex]\(X_c'\)[/tex] and the original capacitive reactance [tex]\(X_c\)[/tex], we can express [tex]\(X_c'\)[/tex] in terms of [tex]\(X_c\)[/tex]:
[tex]\[ X_c' = \frac{1}{4} \times \left( \frac{1}{2 \pi f C} \right) = \frac{X_c}{4} \][/tex]
Thus, the capacitive reactance decreases by a factor of 4 when the frequency is increased to [tex]\(4f\)[/tex]. Hence, the correct statement is:
The capacitive reactance decreases by a factor of four.
