0 0 votes In a series $\mathrm{RLC}$ circuit, $\mathrm{R}=2 \; \mathrm{k} \Omega, \mathrm{L}=1 \; \mathrm{H}$, and $\mathrm{C} \frac{1}{400}=\mu \mathrm{F}$. The resonant frequency is $2 \times 10^{4} \mathrm{~Hz}$ $\frac{1}{\pi} \times 10^{4} \mathrm{~Hz}$ $10^{4} \mathrm{~Hz}$ $2 \pi \times 10^{4} \mathrm{~Hz}$ Circuit Analysis gate2005-ec analog-circuits rlc-circuits frequency-response numerical-answers + – admin 263 views answer comment Share Follow Add Sync Questions Print 0 reply Please log in or register to add a comment.
