Keeping the source of frequency equal to the resonating frequency of the series LCR circuit, if the three elements L, C and R in are arranged in parallel , show that the total current in the parallel LCR circuit is a minimum at this frequency. Obtain the r.m.s. value of current in each branch of the circuit for the elements and source specified in for this frequency.

Explanation

Inductor (L),a capacitor (C),and a resistor (R) is connected in parallel with each other in a circuit where,
L=5.0H
C=80 `muF`=`80xx10^(-6)F`
`R=40Omega`
Potential of the voltage source, `V=230V`
Impedance (Z) of the given parallel LCR circuit is given as:
`(1)/(2)=sqrt((1)/(R^(2))+((1)/(omegaL)-omegaC)^(2))`
where,
`omega=` Angular frequency
At resonance, `(1)/(omegaL)-omegaC=0`
`therefore omega=(1)/sqrt(LC)`
`=(1)/(sqrt(5xx 80xx10^(-6)))=50` rad/s
Hence, the magnitude of Z is the maximum at 50 rad/s . As a result at the total current is minimum.
Rms current flowing through inductor L is given as:
`I_(L)=(V)/(omegaL)`
`=(230)/(50xx5)=0.92A`
Rms current flowing through capacitor C is given as:
`I_(C)=(V)/(A)=omegaCV`
`omegaC`
`=50xx80xx10^(-6)xx230=0.92A`
Rms current flowing through resistor R is given as:
`I_(R)=(V)/(R)`
`=(230)/(4)=5.75A`