Answer
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Hint:Any circuit containing a capacitor, the capacitive reactance offers the resistance to the current flowing through the capacitor. The capacitive reactance is inversely proportional to the frequency of the circuit. As a direct current has zero frequency, the capacitor offers infinite resistance to the current flowing through it.
Complete step by step answer:
To understand clearly, let's first consider an a.c. circuit containing a capacitor of capacitance $C$.The e.m.f. of a.c. source is,
$E = {E_0}\sin \omega t$
Where $\omega $ is the angular frequency of the a.c. source.
We know that the expression to calculate the charge on the capacitor is $q = EC$.
$q = C{E_0}\sin \omega t$
The current flowing through the capacitor is $I = \dfrac{{dq}}{{dt}}$.
$I = \dfrac{d}{{dt}}\left( {C{E_0}\sin \omega t} \right)$
$ \Rightarrow I = \omega C{E_0}\cos \omega t$
We can arrange the above equation as follows
$I = \dfrac{{{E_0}}}{{\left( {\dfrac{1}{{\omega C}}} \right)}}\sin \left( {\omega t + \dfrac{\pi }{2}} \right)$
The peak value of a.c. current,
${I_0} = \dfrac{{{E_0}}}{{\left( {\dfrac{1}{{\omega C}}} \right)}}$
Now it is clear that the factor $\dfrac{1}{{\omega C}}$ offers opposition to the flow of current through the capacitor. This factor is called capacitive reactance ${X_C}$.
${X_C} = \dfrac{1}{{\omega C}}$
Also, we know that $\omega = 2\pi f$.
Where, $f$ is the frequency of the a.c. source.
The capacitive reactance becomes
${X_C} = \dfrac{1}{{2\pi fC}}$
For a.c. circuit $f \ne 0$, therefore a.c. current can pass through a capacitor.
Now comes to the D.C. circuit containing a capacitor only.
For D.C. circuit, the frequency $f = 0$
$\therefore {X_C} = \dfrac{1}{0} = \infty $
Therefore, a capacitor offers infinite opposition to d.c. current.In other words, a capacitor block d.c. current but passes a.c. current.
Note:In d.c. circuit, the polarity of the source does not alternate with time and hence the current flows steadily in one direction. Therefore, the frequency of d.c. circuit is zero.But in a.c. circuit, the magnitude of e.m.f. of the source changes with time and direction reverses periodically. Therefore, an a.c. the circuit has a frequency.
Complete step by step answer:
To understand clearly, let's first consider an a.c. circuit containing a capacitor of capacitance $C$.The e.m.f. of a.c. source is,
$E = {E_0}\sin \omega t$
Where $\omega $ is the angular frequency of the a.c. source.
We know that the expression to calculate the charge on the capacitor is $q = EC$.
$q = C{E_0}\sin \omega t$
The current flowing through the capacitor is $I = \dfrac{{dq}}{{dt}}$.
$I = \dfrac{d}{{dt}}\left( {C{E_0}\sin \omega t} \right)$
$ \Rightarrow I = \omega C{E_0}\cos \omega t$
We can arrange the above equation as follows
$I = \dfrac{{{E_0}}}{{\left( {\dfrac{1}{{\omega C}}} \right)}}\sin \left( {\omega t + \dfrac{\pi }{2}} \right)$
The peak value of a.c. current,
${I_0} = \dfrac{{{E_0}}}{{\left( {\dfrac{1}{{\omega C}}} \right)}}$
Now it is clear that the factor $\dfrac{1}{{\omega C}}$ offers opposition to the flow of current through the capacitor. This factor is called capacitive reactance ${X_C}$.
${X_C} = \dfrac{1}{{\omega C}}$
Also, we know that $\omega = 2\pi f$.
Where, $f$ is the frequency of the a.c. source.
The capacitive reactance becomes
${X_C} = \dfrac{1}{{2\pi fC}}$
For a.c. circuit $f \ne 0$, therefore a.c. current can pass through a capacitor.
Now comes to the D.C. circuit containing a capacitor only.
For D.C. circuit, the frequency $f = 0$
$\therefore {X_C} = \dfrac{1}{0} = \infty $
Therefore, a capacitor offers infinite opposition to d.c. current.In other words, a capacitor block d.c. current but passes a.c. current.
Note:In d.c. circuit, the polarity of the source does not alternate with time and hence the current flows steadily in one direction. Therefore, the frequency of d.c. circuit is zero.But in a.c. circuit, the magnitude of e.m.f. of the source changes with time and direction reverses periodically. Therefore, an a.c. the circuit has a frequency.
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