KCET · Physics · Alternating Current
A series combination of resistor \((R)\), capacitor (C) is connected to an A.C. source of angular frequency \(\omega\). Keeping the voltage same, if the frequency is changed to \(\omega / 3\), the current becomes half of the original current. Then the ratio of the capacitive reactance and resistance at the former frequency is
- A \(\sqrt{0.6}\)
- B \(\sqrt{3}\)
- C \(\sqrt{2}\)
- D \(\sqrt{6}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{0.6}\)
Step-by-step Solution
Detailed explanation
The impedance of the circuit, \(Z=\sqrt{R^{2}+X_{C}^{2}}\)
When angular frequency of source is reduced to \(\frac{1}{3}\), the capacitor reactance is increased by 3 times. \(\therefore Z^{\prime}=\sqrt{R^{2}+\left(3 X_{C}\right)^{2}}=\sqrt{R^{2}+9 X_{C}^{2}}\)
As current becomes half \(\therefore \quad Z^{\prime}=2 Z\)
\(\therefore 2 Z=\sqrt{R^{2}+9 X_{C}^{2}} \)
\( \Rightarrow 2 \sqrt{R^{2}+X_{C}^{2}}=\sqrt{R^{2}+9 X_{C}^{2}} \)
\( \Rightarrow 4\left(R^{2}+X_{C}^{2}\right)=R^{2}+9 X_{C}^{2} \)
\( \Rightarrow 3 R^{2}=5 X_{C}^{2} \)
\( \text {or } \frac{X_{C}}{R}=\sqrt{\frac{3}{5}}=\sqrt{0.6}\)
When angular frequency of source is reduced to \(\frac{1}{3}\), the capacitor reactance is increased by 3 times. \(\therefore Z^{\prime}=\sqrt{R^{2}+\left(3 X_{C}\right)^{2}}=\sqrt{R^{2}+9 X_{C}^{2}}\)
As current becomes half \(\therefore \quad Z^{\prime}=2 Z\)
\(\therefore 2 Z=\sqrt{R^{2}+9 X_{C}^{2}} \)
\( \Rightarrow 2 \sqrt{R^{2}+X_{C}^{2}}=\sqrt{R^{2}+9 X_{C}^{2}} \)
\( \Rightarrow 4\left(R^{2}+X_{C}^{2}\right)=R^{2}+9 X_{C}^{2} \)
\( \Rightarrow 3 R^{2}=5 X_{C}^{2} \)
\( \text {or } \frac{X_{C}}{R}=\sqrt{\frac{3}{5}}=\sqrt{0.6}\)
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