question_answer

Consider an AC circuit as shown in the figure above. If \[{{R}^{2}}=\frac{L}{C}\] which one of the following statements is correct?            

A) Impedance is independent of frequency

B) Impedance is dependent of frequency

C) Impedance is the resistance only

D) None of the above

Correct Answer: C

Solution :

[c]

Here

(i) Rand L is series wire in wire AB so reactance \[{{\Zeta }_{1}}=R+\operatorname{j}\omega L\]

(ii) C and R is series wire. So, \[{{\Zeta }_{2}}=R+\frac{1}{\operatorname{j}\omega L}\]

The equivalent circuit is shown as,

Hence, impedance Z is given as \[\frac{1}{Z}=\frac{1}{{{Z}_{1}}}+\frac{1}{{{Z}_{2}}}\]

\[=\frac{1}{R+j\omega L}+(R+j\omega C)\]\[=\frac{1+(R+j\omega C)(R+j\omega L)}{R+j\omega L}\]\[=\frac{(1+R-{{\omega }^{2}}CL)+j(\omega CR+\omega LR)}{R+j\omega L}\]

\[\Rightarrow \]\[Z=\frac{R+j\omega L}{(1+{{R}^{2}}-{{\omega }^{2}}CL)+j\omega L(C+L)}\]\[=\frac{[R+j\omega L][1+{{R}^{2}}-{{\omega }^{2}}CL-j\left( \omega R \right)\left( C+L \right)]}{(1+{{R}^{2}}-{{\omega }^{2}}CL)+{{\omega }^{2}}{{R}^{2}}{{(C+L)}^{2}}}\]

Hence, the imaginary part of the impedance is gives as \[=\omega L(1+{{R}^{2}}-{{\omega }^{2}}CL)-{{\omega }^{2}}{{R}^{2}}(C+L)\]

Putting \[{{R}^{2}}=\frac{L}{C}\]we get it zero.