JEE Mains · Physics · STD 12 -7. Alternating current
An a.c. source of angular frequency \(\omega\) is connected across a resistor \(R\) and a capacitor \(C\) in series. The current is observed as \(I\). Now the frequency of the source is changed to \(\omega/4\), (keeping the voltage unchanged) the current is found to be \(I/3\). The ratio of resistance to reactance at frequency \(\omega\) is
- A \(\sqrt{\dfrac{6}{7}}\)
- B \(\sqrt{\dfrac{3}{5}}\)
- C \(\sqrt{\dfrac{7}{8}}\)
- D \(\sqrt{\dfrac{3}{4}}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\dfrac{7}{8}}\)
Step-by-step Solution
Detailed explanation
The current in the RC circuit at frequency \(\omega\) is given by \(I = \dfrac{V}{\sqrt{R^2 + X_C^2}}\) where \(X_C = \dfrac{1}{\omega C}\) is the capacitive reactance. When the frequency is changed to \(\omega/4\), the new reactance becomes…
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