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# A $15.0\mu F$ capacitor is connected to $220V,\;50Hz$ source. The capacitive reactance and the RMS current would be: A) $312.31\Omega \\1.03{\text{ }}A \\$ B) $212.31\Omega \\2.03{\text{ }}A \\$C) $312.31\Omega \\2.03{\text{ }}A \\$D) $212.31\Omega \\1.03{\text{ }}A \\$

Last updated date: 13th Jun 2024
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Hint: To solve this question, i.e., to find the values of capacitive reactance and the RMS current, we will apply the formula of both capacitive reactance and the RMS current, as the direct formula of both the values is available. Then on putting the given values in both the formulas, we will get our required answer.

We have been given that a $15.0\mu F$ capacitor is connected to the$220V,\;50Hz$ source. We need to find the capacitive reactance and the RMS current.
So, we have been given the capacity, $C{\text{ }} = {\text{ }}15{\text{ }} \times {\text{ }}{10^{ - 6}}\;F$
The given Rms value of voltage, ${V_{rms}}{\text{ }} = {\text{ }}220\;V$
Also, the frequency of source, $f{\text{ }} = {\text{ }}50\;Hz$
So, the formula of capacitive reactance is, ​${X_c} = \dfrac{1}{{2\pi fC}}$​
On putting the formula in the above formula, we get
$\Rightarrow{X_c} = \dfrac{1}{{2\pi \times 50 \times 15 \times {{10}^{ - 6}}}} \\ \Rightarrow{X_c} = 212.31\Omega \\$
And the formula for RMS value of current is, ​ ${I_{rms}} = \dfrac{{{V_{rms}}}}{{{X_c}}}$
​​On putting the values in the above formula, we get
$\Rightarrow{I_{rms}}{\text{ }} = {\text{ }}\dfrac{{220}}{{212.31}} = 1.03\;A$
So, the capacitive reactance and the RMS current is $212.31\Omega$ and $1.03A$ respectively.

Thus, option (D) $212.31\Omega$ and $1.03{\text{ }}A,$ is correct.

Note: In the question, we are asked about the capacitive reactance and the RMS current, let us understand about them in detail. So, capacitive reactance is a measure of a capacitor's opposition to alternating current. Like resistance it is also measured in ohms, but reactance value depends on the frequency of the signal passing through the capacitor. Its symbol is ${X_c}$. And the RMS current value is the direct current that disintegrates the same power in a resistor.