$\omega_{\text{r}}=\frac{1}{\sqrt{\text{LC}}}$
$=\frac{1}{\sqrt{5\times80\times10^{-6}}}$
$=\frac{1}{2\times10^{-2}}$
= 50rad/sec.
$\text{Z}=\sqrt{\text{R}^2+\big(\omega\text{L}-\frac{1}{\omega\text{C}}\big)^2}$
At resonance, $\omega\text{L}=\frac{1}{\omega\text{C}}$
Therefore,
$\text{Z}=\sqrt{\text{R}^2}=\text{R}=40\Omega$
Amplitude of current at resonating frequency
Peak value of current, $\text{I}_0=\frac{\text{E}_0}{\text{z}}=\frac{\sqrt{2}\times230}{40}=8.13\text{A}$
Rms value of current, $\text{I}_{\text{v}}=\frac{\text{I}_0}{\sqrt{2}}=\frac{8.13}{\sqrt{2}}=5.75\text{A}$
$\text{V}_{\text{L rms}}=\text{I}_\text{v}\omega_{\text{r}}\text{L}$
= 5.75 × 50 × 5.0
= 1437.5V
Potential drop across R
VR rms = Iv × R
= 5.75 × 40
= 230volts
Potential drop across C
$\text{V}_{\text{C rms}}=\text{I}_{\text{v}}\Big(\frac{1}{\omega_{\text{r}}\text{C}}\Big)$
$=5.75\times\frac{1}{50\times80\times10^{-6}}$
$=\frac{5.75}{4}\times10^3$
= 1437.5V
Therefore,
Potential drop across LC circuit
VLC rms = VL rms - VC rms = 0
Thus, the potential drop across the LC combination is zero at the resonating frequency.
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