Keeping the frequency equal to the resonating frequency of the series LCR circuit- the three elements- L- C- and R are arranged in parallel- show that the total current in the parallel LCR circuit is minimum this frequency- Obtain the current RMS value in each branch of the circuit the elements and specified as below this frequency-The figure shows a series LCR circuit connected to a variable frequency 230 V source- L - 5-0 H- C - 80 mu F- R - 40Omega -

Question

Toppr Admin · Accepted Answer

1-Z-x221A-1R2-1-x3C9-L-x2212-x3C9-C-2At resonance- -x3C9-x221A-1-LC-50rad-sat -x3C9-50- Z is maximum and current is minimum-IL-V-x3C9-L-0-92AIc-x3C9-CV-0-92AIR-V-R-5-75A

1/Z=√1R2+(1/ωL−ωC)2At resonance, ω=√1/LC=50rad/sat ω=50, Z is maximum and current is minimum.IL=V/ωL=0.92AIc=ωCV=0.92AIR=V/R=5.75A

$1 / Z = \sqrt{\frac{1}{R^{2}} + (1 / ω L - ω C)^{2}}$
At resonance, $ω = \sqrt{1 / L C} = 50 r a d / s$
at $ω = 50$ , $Z$ is maximum and current is minimum.
$I_{L} = V / ω L = 0.92 A$
$I_{c} = ω C V = 0.92 A$
$I_{R} = V / R = 5.75 A$