MHT CET · Physics · Alternating Current
In a capactive circuit, the reactance of capacitor at frequency 'f' is ' \(\mathrm{X}_{\mathrm{c}}{ }^{\prime}\). What will be
its reactance at frequency \(4 \mathrm{f} ?\)
- A \(\frac{\mathrm{X}_{\mathrm{C}}}{2}\)
- B \(\frac{X_{C}}{4}\)
- C \(\frac{\mathrm{X}_{\mathrm{C}}}{8}\)
- D \(X_{C}\)
Answer & Solution
Correct Answer
(B) \(\frac{X_{C}}{4}\)
Step-by-step Solution
Detailed explanation
Capacitive reactance \(\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \pi \mathrm{fC}}\)
\(\begin{array}{l}
\therefore \frac{\mathrm{X}_{\mathrm{C}}^{\prime}}{\mathrm{X}_{\mathrm{c}}}=\frac{\mathrm{f}}{\mathrm{f}^{\prime}}=\frac{1}{4} \\
\therefore \mathrm{X}_{\mathrm{c}}^{\prime}=\frac{\mathrm{X}_{\mathrm{c}}}{4}
\end{array}\)
\(\begin{array}{l}
\therefore \frac{\mathrm{X}_{\mathrm{C}}^{\prime}}{\mathrm{X}_{\mathrm{c}}}=\frac{\mathrm{f}}{\mathrm{f}^{\prime}}=\frac{1}{4} \\
\therefore \mathrm{X}_{\mathrm{c}}^{\prime}=\frac{\mathrm{X}_{\mathrm{c}}}{4}
\end{array}\)
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