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Q. The impedance of a $ RC $ circuit is $ {{Z}_{1}} $ for frequency $ f $ and $ {{Z}_{2}} $ for frequency $ 2f $ . Then, $ {{Z}_{1}}/{{Z}_{2}} $ is

KEAMKEAM 2009Alternating Current

Solution:

The impedance of R-C circuit for frequency
$ {{f}_{1}} $ is $ {{Z}_{1}}=\sqrt{{{R}^{2}}+\frac{1}{4{{\pi }^{2}}{{f}^{2}}{{C}^{2}}}} $
The impedance of JR-C circuit for frequency $ 2f $ is $ {{Z}_{2}}=\sqrt{{{R}^{2}}+\frac{1}{4{{\pi }^{2}}(2{{f}^{2}}){{C}^{2}}}} $
Or $ {{Z}_{2}}=\sqrt{{{R}^{2}}+\frac{1}{16{{\pi }^{2}}{{f}^{2}}{{C}^{2}}}} $
Then $ \frac{Z_{1}^{2}}{Z_{2}^{2}}=\frac{{{R}^{2}}+\frac{1}{4{{\pi }^{2}}{{f}^{2}}{{C}^{2}}}}{{{R}^{2}}+\frac{1}{16{{\pi }^{2}}{{f}^{2}}{{C}^{2}}}} $
Or $ \frac{Z_{1}^{2}}{Z_{2}^{2}}=\frac{1+\frac{1}{4{{\pi }^{2}}{{f}^{2}}{{C}^{2}}{{R}^{2}}}}{1+\frac{1}{16{{\pi }^{2}}{{f}^{2}}{{R}^{2}}{{C}^{2}}}} $
value is greater than 1.
Hence, $ \frac{{{Z}_{1}}}{{{Z}_{2}}}= $ lies between 1 and 2.