Question 254: AC fundamentals

Question

The reactance of a capacitor is $\mathrm{X}_{\mathrm{C}}$. If the frequency and the capacitance are doubled, then the new reactance will be

Accepted Answer

Given, $\mathrm{X}_{\mathrm{C}}=\mathrm{X} \Omega$
$$ \Rightarrow \quad \frac{1}{2 \pi \mathrm{fC}}=\mathrm{X} \Omega $$
New Capacitance $\mathrm{C}=2 \mathrm{C}$ and new frequency
$$ \mathrm{f}^{\prime}=2 \mathrm{f} $$
$	herefore \quad$ New capacitive reactance $\mathrm{X}_{\mathrm{C}}^{\prime}=\frac{1}{2 \pi(2 \mathrm{f})(2 \mathrm{C})}$
$$ \begin{aligned} & \mathrm{X}_{\mathrm{C}}^{\prime}=\frac{1}{(2 \pi)(4 \mathrm{fC})} \ & \mathrm{X}_{\mathrm{C}}^{\prime}=\frac{1}{4} \mathrm{X}_{\mathrm{C}} \ & \mathrm{X}_{\mathrm{C}}^{\prime}=\frac{\mathrm{X}_{\mathrm{C}}}{4} \Omega \end{aligned} $$

Answer

$\frac{X_C}{2}$

Answer

$\mathrm{X}_{\mathrm{C}}$

Answer

$\frac{X_C}{4}$

Answer

$2 X_C$

AC fundamentals

The reactance of a capacitor is \mathrm{X}_{\mathrm{C}}. If the frequency and the capacitance are doubled, then the new reactance will be

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