Question
Figure shows a series LCR circuit connected to a variable frequency 230V source. L = 5.0H, C = 80μF, R = 40Ω.
  1. Determine the source frequency which drives the circuit in resonance
  2. Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
  3. Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.

Answer

Here, we are given a LCR circuit.
Inductance, $L = 5.0H$
Resistance, $\text{R}=40\Omega$
Capacitance, $C = 80\mu F = 80 \times 10^{-6}F$
Effective voltage, $E_v = 230$volt
$\Rightarrow\ \text{Peak voltage},\ \text{E}_0=\sqrt{2}\text{E}_{\text{v}}=\sqrt{2}\times230\text{V}$
  1. Resonance angular frequency is given by
$\omega_{\text{r}}=\frac{1}{\sqrt{\text{LC}}}$
$=\frac{1}{\sqrt{5\times80\times10^{-6}}}$
$=\frac{1}{2\times10^{-2}}$
= 50rad/sec.
  1. Impedance of the circuit,
$\text{Z}=\sqrt{\text{R}^2+\big(\omega\text{L}-\frac{1}{\omega\text{C}}\big)^2}$
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, $\text{I}_{\text{v}}=\frac{\text{I}_0}{\sqrt{2}}=\frac{8.13}{\sqrt{2}}=5.75\text{A}$
  1. Potential drop across L
$\text{V}_{\text{L rms}}=\text{I}_\text{v}\omega_{\text{r}}\text{L}$
$= 5.75 × 50 × 5.0$
$= 1437.5V$
Potential drop across R
$\mathrm{V_{R\ rms} = I_v \times\ R}$
$= 5.75 × 40$
$= 230$volts
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
$\mathrm{V_{LC\  rms} = V_{L\ rms} - V_{C\ rms} = 0}$
Thus, the potential drop across the LC combination is zero at the resonating frequency.

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