MHT CET · Physics · Alternating Current
The reactance of capacitor at \(50 \mathrm{~Hz}\) is \(5 \Omega\). If the frequency is increased to \(100 \mathrm{~Hz}\), the new reactance is
- A \(5 \Omega\)
- B \(2.5 \Omega\)
- C \(10 \Omega\)
- D \(125 \Omega\)
Answer & Solution
Correct Answer
(C) \(10 \Omega\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
\mathrm{X}_{\mathrm{C}} & =\frac{1}{2 \pi \mathrm{fc}} \\
\therefore \quad \mathrm{C} & =\frac{1}{2 \pi \mathrm{f}(5)} \quad \ldots .\left(\because \mathrm{X}_{\mathrm{C}}=5 \Omega\right)
\end{aligned}\)
New reactance,
\(\mathrm{X}_{\mathrm{C}}^{\prime}=\frac{1}{2 \pi \mathrm{f}^{\prime} \mathrm{C}}=\frac{1}{2 \pi(2 \mathrm{f}) \mathrm{C}}=\frac{\mathrm{X}_{\mathrm{C}}}{2}=\frac{1}{2} \times 5=2.5 \Omega\)
\mathrm{X}_{\mathrm{C}} & =\frac{1}{2 \pi \mathrm{fc}} \\
\therefore \quad \mathrm{C} & =\frac{1}{2 \pi \mathrm{f}(5)} \quad \ldots .\left(\because \mathrm{X}_{\mathrm{C}}=5 \Omega\right)
\end{aligned}\)
New reactance,
\(\mathrm{X}_{\mathrm{C}}^{\prime}=\frac{1}{2 \pi \mathrm{f}^{\prime} \mathrm{C}}=\frac{1}{2 \pi(2 \mathrm{f}) \mathrm{C}}=\frac{\mathrm{X}_{\mathrm{C}}}{2}=\frac{1}{2} \times 5=2.5 \Omega\)
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