2023 Alternating CurrentImpedance of a Series LCR Circuit Medium Question 12 of 60 in Alternating Current

The net impedance of circuit (as shown in figure) will be:

The net impedance of circuit (as shown in figure) will be:

A $25\ \Omega$
B $10\sqrt{2}\ \Omega$
C $15\ \Omega$
D $5\sqrt{5}\ \Omega$

The answer is D.

Solution

$X_L=5\ \Omega$, $X_C=10\ \Omega$, $R=10\ \Omega$ so $Z=\sqrt{125}=5\sqrt{5}\ \Omega$.

$\omega=2\pi f\Rightarrow\omega=100\pi$ $Z=\sqrt{R^2+(X_L-X_C)^2}$ $=\sqrt{10^2+\left(\omega L-\frac{1}{\omega C}\right)^2}$ $=\sqrt{100+(5-10)^2}=\sqrt{125}$ $Z=5\sqrt{5}\ \Omega$

Concept impedance from X_L, X_C and R

Alternating Current in past papers

60 questions from this chapter have appeared across 24 exam years.