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: Phasor Triangle:
Draw the right triangle with horizontal side $R$ and vertical side $X_L-X_C$.

Step 2: Sides:
$X_L=R$, $X_C=R/2$, so the vertical side is $R/2$. The hypotenuse is $\sqrt{R^2+R^2/4}=\dfrac{\sqrt5}{2}R$.

Step 3: Angle:
$\tan\phi=\dfrac{R/2}{R}=\dfrac12$. Option (C).

Final Answer:
Option (C). \[ \boxed{\text{(C) } \frac{\sqrt{5}R}{2},\ \tan^{-1}\left(\frac{1}{2}\right)} \]