Question

Figure shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80 $\mu$F, R = 40$\Omega$.

(a) Determine the source frequency which drives the circuit in resonance.

(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.

(c) 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 1

(a)

The resonance frequency is given by

$\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{5\times 80\times {10}^{-6}}}=50rad/s$

The resonant frquency is 50 rad/s.

(b)

At resonance, $\omega L=1/\omega C$

$\Rightarrow Z=R=40\Omega$

The peak voltage, $V_o=\sqrt{2}V$

$I_o=\sqrt{2}V/Z$

$I_o=8.13A$

(c)

Potential drop across inductor,

$V_{L\left(rms\right)}=I\times {\omega }_{R}L$

$I=I_0/\sqrt{2}=\sqrt{2}V/\sqrt{2}Z=23/40A$

$\Rightarrow V_{rms}=1437.5V$

Potential drop across capacitor,

$(V_c{)}_{rms}=I\times \frac{1}{w_{r}C}=1437.5V$

Potential drop across resistor,

$(V_R{)}_{rms}=I\times R=230V$

Potential drop across the LC combination

$V_{LC}$= $I$($X_L$-$X_C$)

at resonance $X_L$=$X_C$ $\Rightarrow V_{LC}=0V$