State the condition for resonance in a series RLC circuit. A series circuit has \(R=10\ \Omega\), \(L=0.2\) H and \(C=50\ \mu\)F and is fed from a 100 V variable-frequency supply. Find the resonant frequency, the current at resonance, the quality fac…

Answer: \(f_r=50.33\) Hz, \(I_0=10\) A, \(Q=6.32\), BW \(=7.96\) Hz, \(f_1=46.51\) Hz, \(f_2=54.47\) Hz.

Condition: at resonance \(X_L=X_C\), i.e. \(\omega L=\dfrac1{\omega C}\). Then \(Z=R\) (minimum), current is maximum and in phase with the voltage (pf = 1). This gives \(f_r=\dfrac{1}{2\pi\sqrt{LC}}\).

\[f_r=\frac{1}{2\pi\sqrt{0.2\times50\times10^{-6}}}=\frac{1}{2\pi\times3.162\times10^{-3}}=50.33\text{ Hz}.\]

Current at resonance: \(I_0=\dfrac VR=\dfrac{100}{10}=10\) A.

Quality factor: \(Q=\dfrac1R\sqrt{\dfrac LC}=\dfrac1{10}\sqrt{\dfrac{0.2}{50\times10^{-6}}}=\dfrac{63.25}{10}=6.32\). (So \(V_L=V_C=QV=632\) V at resonance.)

Bandwidth: \(BW=f_2-f_1=\dfrac{R}{2\pi L}=\dfrac{10}{2\pi\times0.2}=7.96\) Hz \((=f_r/Q)\).

Half-power frequencies: \(\omega_{1,2}=\mp\dfrac{R}{2L}+\sqrt{\left(\dfrac R{2L}\right)^2+\dfrac1{LC}}=\mp25+\sqrt{625+10^5}=\mp25+317.21\)

\(\omega_1=292.21\) rad/s → \(f_1=46.51\) Hz; \(\omega_2=342.21\) rad/s → \(f_2=54.47\) Hz. (The quick approximation \(f_r\mp BW/2\) gives 46.35 Hz and 54.31 Hz.)