Demonstrative Video
PARALLEL RESONANCE

also called Tank circuit

Y=H(ω)=IVY=1R+j(ωC−1ωL)ωC−1ωL=0ω0=1√LCrad/sY=H(ω)=IVY=1R+j(ωC−1ωL)ωC−1ωL=0ω0=1√LCrad/s
- ω1=ω0√1+(12Q)2−ω02Q,ω2=ω0√1+(12Q)2+ω02Qω1=ω0√1+(12Q)2−ω02Q,ω2=ω0√1+(12Q)2+ω02QHalf-power frequencies
- ω1≃ω0−B2,ω2≃ω0+B2ω1≃ω0−B2,ω2≃ω0+B2circuits For high
Resonant RLC circuits
Characteristic | Series circuit | Parallel circuit |
---|---|---|
Resonant freq, ω0ω0 | 1√LC1√LC | 1√LC1√LC |
Quality factor, Q | ω0LR or 1ω0RCω0LR or 1ω0RC | Rω0L or ω0RCRω0L or ω0RC |
Bandwidth, B | ω0Qω0Q | ω0Qω0Q |
Half-power freq., ω1,2ω1,2 | ω0√1+(12Q)2±ω02Qω0√1+(12Q)2±ω02Q | ω0√1+(12Q)2±ω02Qω0√1+(12Q)2±ω02Q |
For Q≥10,ω1,ω2Q≥10,ω1,ω2 | ω0±B2ω0±B2 | ω0±B2ω0±B2 |
Problem
Let R=8 kΩR=8 kΩ, L=0.2 mHL=0.2 mH, and C=8 μFC=8 μF. calculate
ω0,1,2, Q, Bω0,1,2, Q, B
power dissipated at ω0,1,2ω0,1,2

ω0=1√LC=25 krad/sQ=Rω0L=1600B=ω0Q=15.625 rad/sω1=ω0−B2=24,992 rad/sω2=ω0+B2=25,008 rad/sω0=1√LC=25 krad/sQ=Rω0L=1600B=ω0Q=15.625 rad/sω1=ω0−B2=24,992 rad/sω2=ω0+B2=25,008 rad/s