Question 375: AC fundamentals

Question

The reactance of capacitor at $$50 \mathrm{~Hz}$$ is $$5 \Omega$$. If the frequency is increased to $$100 \mathrm{~Hz}$$, the new reactance is

Accepted Answer

$$\begin{aligned}
X_C & =\frac{1}{2 \pi f c} \
	herefore \quad C & =\frac{1}{2 \pi f(5)} \quad \ldots .\left(\because X_C=5 \Omegaight)
\end{aligned}$$
New reactance,
$$X_C^{\prime}=\frac{1}{2 \pi f^{\prime} C}=\frac{1}{2 \pi(2 f) C}=\frac{X_C}{2}=\frac{1}{2} 	imes 5=2.5 \Omega$$

Answer

$$5 \Omega$$

Answer

$$10 \Omega$$

Answer

$$2.5 \Omega$$

Answer

$$125 \Omega$$

AC fundamentals

The reactance of capacitor at $50 \mathrm{~Hz} is 5 \Omega. If the frequency is increased to 100 \mathrm{~Hz}$, the new reactance is

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