
For high frequency, capacitor offers
A. High reactance
B. Low reactance
C. Cannot be predicted
D. Infinite reactance
Answer
543.9k+ views
Hint:To solve this question, we need to understand the concept of capacitive reactance. Capacitive reactance can be defined as the opposition offered by a capacitor to the flow of ac current in the ac circuit. A capacitor opposes the changes in the potential difference or the voltage across its plates. We will use the relation between capacitive reactance and angular frequency to determine the answer.
Complete answer:
We know that the capacitive reactance is given by the formula:
\[{X_c} = \dfrac{1}{{\omega c}}\],
where, \[{X_c}\] is the capacitive reactance, \[\omega \] is the angular frequency and \[c\] is the capacitance.
But we know the relation between angular frequency and frequency is given by,
\[\omega = 2\pi f\]
where \[f\]is the frequency.
Now, we will put this value in the formula for capacitive reactance.
\[ {X_c} = \dfrac{1}{{2\pi fc}}\]
For a particular capacitor, \[2\pi c\] is constant.Therefore, we can say that ${X_c} \propto \dfrac{1}{f}$ . This means that the capacitive reactance is inversely proportional to the frequency. So we can say that the value of reactance increases with decrease in frequency and it decreases with increase in frequency.Thus, we can say that for the higher frequency, the capacitor offers low reactance.
Hence, option B is the right answer.
Note:Here, we have seen that capacitor offers lower reactance for higher frequency. However, this is only true for ac supply. This is because in the dc supply, there is no frequency which means the value of frequency is zero in dc supply. Therefore, the capacitive reactance will be infinite for dc supply.
Complete answer:
We know that the capacitive reactance is given by the formula:
\[{X_c} = \dfrac{1}{{\omega c}}\],
where, \[{X_c}\] is the capacitive reactance, \[\omega \] is the angular frequency and \[c\] is the capacitance.
But we know the relation between angular frequency and frequency is given by,
\[\omega = 2\pi f\]
where \[f\]is the frequency.
Now, we will put this value in the formula for capacitive reactance.
\[ {X_c} = \dfrac{1}{{2\pi fc}}\]
For a particular capacitor, \[2\pi c\] is constant.Therefore, we can say that ${X_c} \propto \dfrac{1}{f}$ . This means that the capacitive reactance is inversely proportional to the frequency. So we can say that the value of reactance increases with decrease in frequency and it decreases with increase in frequency.Thus, we can say that for the higher frequency, the capacitor offers low reactance.
Hence, option B is the right answer.
Note:Here, we have seen that capacitor offers lower reactance for higher frequency. However, this is only true for ac supply. This is because in the dc supply, there is no frequency which means the value of frequency is zero in dc supply. Therefore, the capacitive reactance will be infinite for dc supply.
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