
The capacitive reactance in an A.C circuit is:
A. effective resistance due to capacity
B. effective wattage
C. effective voltage
D. none of the above
Answer
553.8k+ views
Hint: In case of DC circuit if there is resistor we can measure using an ohmmeter but if there are capacitors in the AC circuit we can’t measure the obstruction for the flow of current hence there comes a term capacitive reactance and it depends on source angular frequency too.
Formula used:
$\eqalign{
& V = {V_0}\sin (\omega t) \cr
& {X_C} = \dfrac{1}{{\omega C}} \cr
& q = CV \cr
& \dfrac{{dq}}{{dt}} = i = C\dfrac{{dV}}{{dt}} \cr} $
Complete answer:
In capacitors the charge stored inside the capacitor at any time is given as $q = CV$
Since voltage varies with respect to time in AC circuits, charge also varies with respect to time and when charge varies with respect to time then automatically differentiation of charge with respect to time gives us current. If a capacitor is connected to a DC source then voltage is constant hence after a certain time charge on the plates of the capacitor also becomes constant. But in case of AC circuit voltage of AC source varies with time. Hence charge on the capacitor also varies with time and same with the current
Normally AC voltages will be in the form $V = {V_0}\sin (\omega t)$ where ${V_0}$ is the peak voltage while $\omega $ is the angular frequency.
$\eqalign{
& q = CV \cr
& \dfrac{{dq}}{{dt}} = i = C\dfrac{{dV}}{{dt}} \cr} $
Now as the rate of change of voltage is more current will be more which is clearly visible from the formula.
Similar to resistance here we use the term reactance for capacitor which will be given as ${X_C} = \dfrac{1}{{\omega C}}$
Where C is the capacitance of the capacitor. And this depends on the source angular frequency too.
Both resistance and capacitive reactance will have a unit as OHM.
So, the correct answer is “Option A”.
Note:
If we consider an sinusoidal AC source we have both positive and negative wave forms for the voltage function hence the charge on the plates of the capacitor will keep on inter changing and in the capacitor dominant circuit, current will be leading the voltage in phase.
Formula used:
$\eqalign{
& V = {V_0}\sin (\omega t) \cr
& {X_C} = \dfrac{1}{{\omega C}} \cr
& q = CV \cr
& \dfrac{{dq}}{{dt}} = i = C\dfrac{{dV}}{{dt}} \cr} $
Complete answer:
In capacitors the charge stored inside the capacitor at any time is given as $q = CV$
Since voltage varies with respect to time in AC circuits, charge also varies with respect to time and when charge varies with respect to time then automatically differentiation of charge with respect to time gives us current. If a capacitor is connected to a DC source then voltage is constant hence after a certain time charge on the plates of the capacitor also becomes constant. But in case of AC circuit voltage of AC source varies with time. Hence charge on the capacitor also varies with time and same with the current
Normally AC voltages will be in the form $V = {V_0}\sin (\omega t)$ where ${V_0}$ is the peak voltage while $\omega $ is the angular frequency.
$\eqalign{
& q = CV \cr
& \dfrac{{dq}}{{dt}} = i = C\dfrac{{dV}}{{dt}} \cr} $
Now as the rate of change of voltage is more current will be more which is clearly visible from the formula.
Similar to resistance here we use the term reactance for capacitor which will be given as ${X_C} = \dfrac{1}{{\omega C}}$
Where C is the capacitance of the capacitor. And this depends on the source angular frequency too.
Both resistance and capacitive reactance will have a unit as OHM.
So, the correct answer is “Option A”.
Note:
If we consider an sinusoidal AC source we have both positive and negative wave forms for the voltage function hence the charge on the plates of the capacitor will keep on inter changing and in the capacitor dominant circuit, current will be leading the voltage in phase.
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