A series circuit consists of a 20 Ω resistor, a 0.1 H inductor and a 50 μF capacitor connected across a 230 V, 50 Hz supply. Calculate (i) the impedance, (ii) the current and the expression for \(i(t)\), (iii) the power factor, (iv) active, reactive…

Answer: \(Z=37.94\ \Omega\) (capacitive), \(I=6.06\) A, \(i=8.57\sin(314.16t+58.19^\circ)\) A, pf \(=0.527\) leading, \(P=734.8\) W, \(Q=1184.8\) VAR (capacitive), \(S=1394.1\) VA, \(V_R=121.2\) V, \(V_L=190.4\) V, \(V_C=385.9\) V, \(f_r=71.18\) Hz.

Reactances: \(X_L=2\pi fL=2\pi(50)(0.1)=31.42\ \Omega\); \(X_C=\dfrac{1}{2\pi fC}=\dfrac{1}{2\pi(50)(50\times10^{-6})}=63.66\ \Omega\).

(i) Impedance: \(X=X_L-X_C=-32.25\ \Omega\) (net capacitive).

\[Z=\sqrt{R^2+X^2}=\sqrt{20^2+32.25^2}=37.94\ \Omega,\quad \phi=\tan^{-1}\frac{-32.25}{20}=-58.19^\circ.\]

(ii) Current: \(I=\dfrac{V}{Z}=\dfrac{230}{37.94}=6.062\) A; \(I_m=\sqrt2\times6.062=8.57\) A.

Taking \(v=325.27\sin(314.16t)\) V, the current leads by 58.19°: \[i(t)=8.57\sin(314.16t+58.19^\circ)\ \text{A}.\]

(iii) Power factor: \(\cos\phi=\dfrac RZ=\dfrac{20}{37.94}=0.527\) leading.

(iv) Powers: \(P=I^2R=6.062^2\times20=734.8\) W; \(Q=I^2X=6.062^2\times32.25=1184.8\) VAR (capacitive); \(S=VI=230\times6.062=1394.1\) VA. Check: \(\sqrt{734.8^2+1184.8^2}=1394.1\) ✓.

(v) Element voltages: \(V_R=IR=121.2\) V, \(V_L=IX_L=190.4\) V, \(V_C=IX_C=385.9\) V. (\(V_C\) is more than the supply voltage; this is possible in a series RLC circuit because \(V_L\) and \(V_C\) are in phase opposition: \(\sqrt{V_R^2+(V_L-V_C)^2}=\sqrt{121.2^2+195.5^2}=230\) V ✓.)

(vi) Resonant frequency: \[f_r=\frac{1}{2\pi\sqrt{LC}}=\frac{1}{2\pi\sqrt{0.1\times50\times10^{-6}}}=71.18\text{ Hz}.\]