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**Hint:**In this question, we need to explain the reason behind the capacitor blocks DC (direct current) and allowing AC (alternating current). We can say that the DC is a fixed value, which means that its polarity (direction) and magnitude do not alter with frequency, whereas AC's polarity and magnitude do.

**Formula used:**

The capacitive reactance is given by

\[{X_C} = \dfrac{1}{{\omega c}} = \dfrac{1}{{2\pi fc}}\]

Here, \[{X_C}\] is capacitive reactance, \[\omega \] is angular frequency, \[c\] is capacitance and \[f\] is frequency of AC supply.

**Complete step by step solution:**

We can say that at first, a capacitor acts as a short circuit, and a fully loaded capacitor acts as an open circuit. Capacitors prevent voltage changes, whereas inductors prevent current changes as well as behave like a DC short circuit. Whenever the capacitor is connected to a DC supply voltage, the positive end of the DC supply at first tries to pull electrons from one terminal and tries to push electrons to the other.

This procedure is iterated until one plate is positively charged while another is negatively charged. The plates of the capacitors are overwhelmed at this point, and no current will pass.The capacitor is now acting as an open circuit. The capacitor will now be destroyed if the DC voltage is increased. If an alternating current voltage is applied to the capacitor, the plates are initially charged. The capacitor will exhaust afterward whenever the direction of the AC supplies changes. Because of the alteration in supply voltage, this process repeats.

We know that capacitive reactance is \[{X_C} = \dfrac{1}{{\omega c}} = \dfrac{1}{{2\pi fc}}\]

Now, put \[f = 0\] in the equation of capacitive reactance.

Thus, we get \[{X_C} = \dfrac{1}{0} = \infty \]

That means a capacitor resists the flow of current as it offers huge reactance. So, there is no flow of current. While in case of AC, the frequency is finite hence the value of reactance is also finite thus allowing the AC current to flow through it.

In that case current in capacitive circuit is \[I = \dfrac{V}{{{X_C}}}\]

Now, put \[{X_C} = \infty \] in the above equation.

Thus, we can say that the current value is zero.

Hence, the capacitor acts as a block for DC and gives a path to AC.

**Therefore, the capacitor blocks DC and allows AC./b>**

**Note:**Here, students generally explain this with the help of theoretical background. But it is also necessary to explain with the help of an equation of capacitive reactance. Moreover the concept of variation of capacitive reactance with rest to frequency is important.**Recently Updated Pages**

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