MHT CET · Physics · Alternating Current
At a particular angular frequency, the reactance of capacitor and that of inductor is same. If the angular frequency is doubled, the ratio of the reactance of the capacitor to that of the inductor will be
- A \(\frac{1}{4}\)
- B \(\frac{1}{2}\)
- C 2
- D 4
Answer & Solution
Correct Answer
(A) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
We know,
\(X_L=\omega L\) ....(i)
\(\Rightarrow X_L \propto \omega\)
Similarly,
\(\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega}\)
Given \(X_L^1=2 X_L\) \(\ldots .(\because \omega=2 \omega)\)
\(X_C^1=\frac{X_c}{2}\)
\(\therefore \quad \frac{\mathrm{X}_{\mathrm{C}}^1}{\mathrm{X}_{\mathrm{L}}^1}=\frac{\frac{\mathrm{X}_{\mathrm{C}}}{2}}{2 \mathrm{X}_{\mathrm{L}}}=\frac{1}{4}\)
....(from (i))
\(X_L=\omega L\) ....(i)
\(\Rightarrow X_L \propto \omega\)
Similarly,
\(\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega}\)
Given \(X_L^1=2 X_L\) \(\ldots .(\because \omega=2 \omega)\)
\(X_C^1=\frac{X_c}{2}\)
\(\therefore \quad \frac{\mathrm{X}_{\mathrm{C}}^1}{\mathrm{X}_{\mathrm{L}}^1}=\frac{\frac{\mathrm{X}_{\mathrm{C}}}{2}}{2 \mathrm{X}_{\mathrm{L}}}=\frac{1}{4}\)
....(from (i))
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