Question

Q16. A15°F capacitor is connected
a 220V, 50Hz a.c. supply. Its
capacitive reactance is​

Answer 1

AnswEr :

We're given with a 15 µF capacitor is connected to a 220 V, 50 Hz A.C. supply. And we've to calculate it's capacitive reactance.

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Now, to calculate capacitive reactance we know the required formula, that is,

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$\: \: \: \: \: \dag \: { \underline{ \boxed{ \sf{ \purple {\: \: X_c = \dfrac{1}{2\pi fC}}}}}}$

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Where, Xc is the capacitive reactance, f is the frequency and C is the given capacitance.

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So, we've the given information, that is,

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• Frequency (f) = 50 Hz

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• Capacitance (C) = 15 µF = 15 x 10⁻⁶ F [Conversion]

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Now, let's substitute the given values in the above formula.

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$:\implies \sf \: \: X_c = \dfrac{1}{2\pi fC}$

$:\implies \sf \: \: X_c = \dfrac{1}{2\pi \times 50 \times 15 \times {10}^{ - 6} }$

$:\implies \sf \: \: X_c = \dfrac{1}{100\pi \times 15 \times {10}^{ - 6} }$

$:\implies \sf \: \: X_c = \dfrac{1}{1500\pi \times {10}^{ - 6} }$

$:\implies \sf \: \: X_c = \dfrac{1}{1500 \times 3.14 \times {10}^{ - 6} }$

$:\implies \sf \: \: X_c = \dfrac{1}{4710 \times {10}^{ - 6} }$

$:\implies \sf \: \: X_c = \dfrac{1}{4710 } \times \dfrac{1}{{10}^{ - 6}} \: \: \: \: \: \: \: \: \: \: \: \bigg(rewriting \: the \: eqn)$

$:\implies \sf \: \: X_c = \bigg(\dfrac{0.0002123}{{10}^{ - 6}} \bigg)$

$:\implies \sf \: \: { \underline{ \boxed{ \frak{ \red{X_c = 212.3 \: \Omega}}}}}$

Hence, capacitive reactance is 212.3 Ω.