Question

The figure show a series LCR circuit with $L=5.0H,C=80\mu F,R=40\Omega$ connected to variable frequency 240 V source. Calculate
(i)The angular frequency of the source which driver the circuit at resonance
(ii) The current at the frequency.
(iii)The rms potential drop across the capacitor at resonance

Answer 1

Solution:-

Given $L=5HC=80\times {10}^{-6}FR=40\Omega$

$V_{eff.}=240V,V_{peak}=\sqrt{2}\times 240V$

a) Resonance angular frequency is

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

b) At resonance

Peak value of current $\Rightarrow I_0=\frac{V_{peak}}{R}=\frac{\sqrt{2}\times 240}{40}=8.48A$

RMS value of current $\Rightarrow I_V=\frac{I_{peak}}{\sqrt{2}}=\frac{8.48}{\sqrt{2}}=5.99A$

c) Potential drop across C:-

$V_{C\left(rms\right)}=I_{rms}\left(\frac{1}{{\omega }^{2}C}\right)=5.99[\frac{{10}^6}{50\times 80}=1497.5V$