Topic: Network Theory
The values of R, L and C in series RLC circuit that resonates at 1.5 kHz and consumes 50 W from a 50 V ac source operating at the resonant frequency. The bandwidth is 0.75 kHz.
- A. $R=50~ohm$, $L=10.6~mH$, $C=1.06~\mu F.$
- B. $R=500~ohm$, $L=10.6~mH$, $C=10.6~\mu F.$
- C. $R=50~ohm$, $L=1.06~mH,$ $C=10.6~\mu F.$
- D. $R=500~ohm$, $L=1.06~mH$, $C=1.06~\mu F$
Answer: A. $R=50~ohm$, $L=10.6~mH$, $C=1.06~\mu F.$
Explanation: At resonance RLC circuit impedance is resistive. $R=\frac{V^{2}}{P}=\frac{50\times50}{50}=50~\Omega$. For series RLC circuit, $B.W.=\frac{R}{2\pi L} \Rightarrow L=\frac{R}{2\pi\times B.W.} = \frac{50}{2\pi\times0.75\times10^{3}}=10.6~mH$. Also, $f_{0}=\frac{1}{2\pi\sqrt{LC}} \Rightarrow (f_{0})^{2}=\frac{1}{4\pi^{2}LC} \Rightarrow C=\frac{1}{4\pi^{2}\times L\times(f_{0})^{2}} = \frac{1}{4\pi^{2}\times10.6\times10^{-3}\times(1.5\times10^{3})^{2}} = 1.06~\mu F$.