A series LCR circuit connected to a variable frequency 230 V source has L = 5.0 H, `C = 80 mu F`, `R = 40 Omega`, Fig.
(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and amplitude of current at at resonating frequency.
(c ) Determine the r.m.s. potential drops acorss the three elements the circuits. Show that the potential drop acorss the LC combination is zero at the resonating frequecny.

Explanation

Inductance of the inductor,L=5.0H
Capacitance of the capacitor, C=80`muH=80xx10^(-6)F`
Resistance of the resistor, R=`40Omega`
Potential of the variable voltage source, V=230V
(a) Resonance angular frequency is given as:
`omega_(R)=(1)/sqrt(LC)`
`=(1)/sqrt(5xx80xx10^(-6))=(10^(3))/(20)=`50rad/s
Hence, the circuit will come in resonance for a source frequency of 50 rad/s.
(b) Impedanc of the circuit is given by the relation,
`Z=sqrt(R^(2)+(omegaL-(1)/(omegaC))^(2))`
At resonance,
`omegaL=(1)/(omegaC)`
`therefore Z=R40Omega`
Amplitude of the current at the resonating frequency is given as `I_(0)=(V_(0))/( Z)`
where,
`v_(0)=` Peak voltage
`=sqrt(2V)`
`therefore I_(0)=sqrt(2V)/(Z)`
`=(sqrt(2)xx230)/(40)=8.13A`
Hence, at resonance, the impedance of the circuit is 40 `Omega` and the amplitude of the current is 8.13A.
(c) RMS potential drop cross the inductor,
`(V_(L))_(rms)=Ixxomega_(R)L`
Whre,
I=rms current
`=(I_(0))/sqrt(2)=sqrt(2V)/(sqrt(2)Z)=(230)/(40)A`
`therefore (V_(L))_(rms)=(230)/(40)xx50xx5=1437.5V`
Potential drop across the capacitor,
`V_(c))_(Rms)=lxx(L)/(omega_(R)C)`
`=(230)/(40)xx(1)/(50xx80xx10^(-6))=1437.5V`
Potential drop across the resistor.
`(V_(R))_(rms)=IR`
`(230)/(40)xx40=230V`
Potential drop across the LC combination,
`V_(LC)=I(omega_(R)L-(1)/(omega_(R)C))` ltbgt At resonance, `(omega _(R)L-(1)/(omega_(R)C))`
`thereforeV_(LC)=0`
Hence, it is proved that the potential drop across the LC combination is zero zt resonating frequency.