A)
B)
C)
D)
Correct Answer: C
Solution :
\[Z=\sqrt{{{R}^{2}}+{{\left( 2\pi fL-\frac{1}{2\pi fC} \right)}^{2}}}\] From above equation at f = 0 \[\Rightarrow z=\infty \] When \[f=\frac{1}{2\pi \sqrt{LC}}\] (resonant frequency) \[\Rightarrow Z=R\] For \[f>\frac{1}{2\pi \sqrt{LC}}\Rightarrow \]Z starts increasing. i.e., for frequency 0 ? fr, Z decreases and for fr to ¥, Z increases. This is justified by graph c.
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