Parallel Resonant Circuit

Solution

For a parallel rlc circuit admittance offered by the circuit Y=G+jB=G+j(B_C–B_L)

Here G= conductance of resistor R

capacitive susceptance B_C=ωC

inductive susceptance B_L=1ωL

For a parallel resonant circuit admittance offered by the circuit Y=G

That means in parallel resonant circuit B_C=B_L

ω_rC=1/ω_rL

1. So, resonant frequency ω_r=1/√L.C=1/√(0.4×10^-6)(16×10^-3)=12.5 ×10³ rad/sec

And f_r=ω_r/2π=1989.4 Hz≈1989 Hz

2. Voltage across the circuit at resonance V=I.R

Here, from the circuit given in the problem I=3.6 mA∠0°

And resistance R=500

So, V=I.R=3.6×10^-3∠0°.500=1.8∠0° V

3. Current through resistor I_R=V/R=1.8∠0°/500=3.6∠0° mA

and current through inductor I_L=V(-J_BL)

here B_L=1/ωL=1/12.5×10^-3(16×10^-9)=1200 siemens

I_L=V(-J_BL)=1.8∠0°×1/200-90°=9 mA∠-90°

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