
The value of capacitive reactance $ ({X_C}) $ for a capacitor having a constant potential difference $ (V) $ between its plates $ \left( {V \ne 0} \right) $ will be
A) zero
B) $ - 14 $
C) $ - 1 $
D) infinite
Answer
474k+ views
Hint: We need to use the relation of a capacitive reactance with the frequency of the potential. Since the potential is constant, we can assume the frequency of the potential as zero and then we can calculate the capacitive reactance.
Formula used: In this solution, we’re going use to use the following formula:
$\Rightarrow {X_C} = \dfrac{1}{{2\pi fC}} $ where $ f $ is the frequency of the varying potential and $ C $ is the capacitance of the capacitor.
Complete step by step solution:
The capacitive reactance for a capacitor in an AC circuit is written as
$\Rightarrow {X_C} = \dfrac{1}{{2\pi fC}} $
However, we’ve been given that the potential applied to the capacitor plates is constant and has a value $ (V) $ . Since the potential is not varying in time as it does in an AC circuit, we can say that the frequency of the potential applied to the capacitor plates is zero i.e. $ f = 0 $ . Substituting the value of $ f = 0 $ in $ {X_C} = \dfrac{1}{{2\pi fC}} $ , we can calculate the capacitive reactance of the circuit as
$\Rightarrow {X_C} = \dfrac{1}{{2\pi (0)C}} $
Since the zero is in the denominator, the capacitive reactance of the capacitor is infinite $ {X_C} = \infty $ which corresponds to option (D).
Note:
While we have calculated the capacitive reactance, in this case, the term is only used for AC circuits where the potential applied to the plates of the capacitor is varying with time, in usually a sinusoidal manner. For very frequently varying potentials i.e. $ V \to 0 $ also, the capacitive reactance $ {X_C} \to \infty $ since the capacitor will allow all the charge to flow in both directions without any resistance as it won't be charged in one either cycle significantly enough.
Formula used: In this solution, we’re going use to use the following formula:
$\Rightarrow {X_C} = \dfrac{1}{{2\pi fC}} $ where $ f $ is the frequency of the varying potential and $ C $ is the capacitance of the capacitor.
Complete step by step solution:
The capacitive reactance for a capacitor in an AC circuit is written as
$\Rightarrow {X_C} = \dfrac{1}{{2\pi fC}} $
However, we’ve been given that the potential applied to the capacitor plates is constant and has a value $ (V) $ . Since the potential is not varying in time as it does in an AC circuit, we can say that the frequency of the potential applied to the capacitor plates is zero i.e. $ f = 0 $ . Substituting the value of $ f = 0 $ in $ {X_C} = \dfrac{1}{{2\pi fC}} $ , we can calculate the capacitive reactance of the circuit as
$\Rightarrow {X_C} = \dfrac{1}{{2\pi (0)C}} $
Since the zero is in the denominator, the capacitive reactance of the capacitor is infinite $ {X_C} = \infty $ which corresponds to option (D).
Note:
While we have calculated the capacitive reactance, in this case, the term is only used for AC circuits where the potential applied to the plates of the capacitor is varying with time, in usually a sinusoidal manner. For very frequently varying potentials i.e. $ V \to 0 $ also, the capacitive reactance $ {X_C} \to \infty $ since the capacitor will allow all the charge to flow in both directions without any resistance as it won't be charged in one either cycle significantly enough.
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