Question

For a series LCR circuit inductive reactance \(X_L\) is equal to resistance \(R\) and also equal to twice the capacitive reactance \(X_C\). The impedance of the circuit and the phase difference between voltage V and current i are respectively

• A. \(\sqrt{5}R, tan^{-1}(\frac{1}{2})\)
• B. \(\sqrt{5}R, tan^{-1}(2)\)
• C. \(\frac{\sqrt{5}R}{2}, tan^{-1}(\frac{1}{2})\)
• D. \(\frac{\sqrt{5}R}{2}, tan^{-1}(2)\)

Hint:

Net reactance is X_L minus X_C; impedance is the root of R squared plus X squared.

Answer 1

The Correct Option is C

Step 1: Reactances:
Let \(R=X_L\) and \(X_C=\dfrac{X_L}2=\dfrac R2\). The net reactance is
\[ X=X_L-X_C=R-\frac R2=\frac R2 \]

Step 2: Impedance:
\[ Z=\sqrt{R^2+X^2}=\sqrt{R^2+\frac{R^2}4}=\frac{\sqrt5R}2 \]

Step 3: Phase Difference:
\[ \tan\phi=\frac{X_L-X_C}R=\frac{R/2}R=\frac12\Rightarrow\phi=\tan^{-1}\left(\frac12\right) \]

Step 4: Check the Other Options:
Options (A) and (B) have \(Z=\sqrt5R\), which would be right if \(X=2R\). Option (D) has the right impedance but the phase \(\tan^{-1}2\), which would need \(X=2R\).

Final Answer:
\(Z=\dfrac{\sqrt5R}2\) and the phase is \(\tan^{-1}(1/2)\), option (C). \[ \boxed{\text{(C) } \frac{\sqrt{5}R}{2},\ \tan^{-1}\left(\frac{1}{2}\right)} \]