AP EAMCET · PHYSICS · Alternating Current
An AC source of angular frequency \(\omega\) is fed across a resistor \(R\) and a capacitor \(C\) in series. The current flowing in the circuit found to be \(I\). Now the frequency of the source is changed to \(\frac{\omega}{3}\), (maintaining the same voltage) the current in the circuit is found to be halved. What is the ratio of reactance to resistance at the original frequency?
- A \(\sqrt{\frac{5}{7}}\)
- B \(\sqrt{\frac{3}{4}}\)
- C \(\sqrt{\frac{3}{5}}\)
- D \(\sqrt{\frac{7}{5}}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\frac{3}{5}}\)
Step-by-step Solution
Detailed explanation
At angular frequency \(\omega\), current through resistance R and capacitance C is given by \(I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+X_C^2}}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}}\)...(i) When angular frequency is changed to…
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