If E = 200,V,R = 25,Omega,L = 2,H,and C = 2,m | vedclass

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(D) In a series $LCR$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$,where $X_L = 2\pi fL$ and $X_C = \frac{1}{2\pi fC}$.At $f

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$0\,A, 0\,A$

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(D) In a series $LCR$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$,where $X_L = 2\pi fL$ and $X_C = \frac{1}{2\pi fC}$.At $f = 0$ ($DC$ condition),the capacitive reactance $X_C = \frac{1}{2\pi fC} = \infty$. Since $X_C$ is in series,the total impedance $Z = \infty$. Therefore,the current $I = \frac{E}{Z} = \frac{200}{\infty} = 0\,A$.At $f = \infty$,the inductive reactance $X_L = 2\pi fL = \infty$. Since $X_L$ is in series,the total impedance $Z = \infty$. Therefore,the current $I = \frac{E}{Z} = \frac{200}{\infty} = 0\,A$.Thus,the current at both $f = 0$ and $f = \infty$ is $0\,A$.

If $E = 200\,V$,$R = 25\,\Omega$,$L = 2\,H$,and $C = 2\,\mu F$ and the frequency is variable,then the current at $f = 0$ and $f = \infty$ will be respectively:

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