Given that the Inductance of the inductor in the circuit is, L = 5.0 H

Given that the Capacitance of the capacitor in the circuit is, C = 80 μH = 80 × 10 ^{– 6} F

Given that Resistance of the resistor in the circuit, R = 40 Ω

Value of Potential of the variable voltage supply, V = 230 V

(a) Resonance angular frequency is given as:

${\omega }_r=\frac{1}{\sqrt{LC}}{\omega }_r=\frac{1}{\sqrt{5x80x10-6}}{\omega }_r=\frac{{10}^3}{20}=50rad/sec$

Hence, the circuit Will come in resonance for a source frequency of 50 rad/s.

(b) Impedance of the circuit is given by the relation:

$Z=\sqrt{R^2+(X_L-X_C{)}^2}$

At resonance, $X_L=X_C\Rightarrow Z=R=40\mathrm{\Omega }$

The amplitude of the current at the resonating frequency is given as:

Where, ${\mathrm{V}}_{\mathrm{o}}$ = Peak voltage = $\sqrt{2}v$

$\therefore {\mathrm{I}}_0=\frac{\sqrt{2\mathrm{V}}}{\mathrm{Z}}=\frac{\sqrt{2}\times 230}{40}=8.13\mathrm{A}$

Hence, at resonance, the impedance of the circuit is 40 $\mathrm{\Omega }$ and the amplitude of the current is 8.13 A.

(c) rms potential drop across the inductor,

                 ( V _L ) _rms = I x ω _r L

Where

$\begin{array}{l}{\mathrm{I}}_{\mathrm{rms}}=\frac{{\mathrm{I}}_{\mathrm{o}}}{\sqrt{2}}=\frac{\sqrt{2}\mathrm{V}}{\sqrt{2}\mathrm{Z}}=\frac{230}{40}=\frac{23}{4}\mathrm{A}\\ \therefore ({\mathrm{V}}_{\mathrm{L}}{)}_{\mathrm{rms}}=\frac{23}{4}\times 50\times 5=1437.5\mathrm{V}\end{array}$

The potential drop across the capacitor:

$\therefore (V_C{)}_{rms}=I\times \frac{1}{{\omega }_{r}C}=\frac{23}{4}\times \frac{1}{50\times 80\times {10}^{-6}}=1437.5V$

The potential drop across the resistor:

$(V_R{)}_{rms}=IR=\frac{23}{4}\times 40=230V$

The potential drop across the LC combination:
$V_{LC}=I(X_L-X_C)$
At resonance, $X_L=X_C\Rightarrow V_{LC}=0$
Hence, it is proved that the potential drop across the LC combination is zero at resonating frequency.