There is a circuit with an inductor and two capacitors in series. Each capacitor is adjacent to the inductor. Capacitor 1 has charge Q1 and Capacitor 2 has charge Q2. Current is moving in the clockwise direction check this figure
We are asked to derive a formula for the resonant frequency.
1
Expert's answer
2020-04-23T12:45:43-0400
Solution: In the figure, two capacitors are connected in series and in series with inductance. The impedance of the circuit with alternating current frequency w can be written as.
Resonance in a sequential circuit is achieved at a frequency at which ∣Z(w)∣→min. In our case of the absence of resistive elements the exact condition must be met ∣Z(w)∣=0, that is
(3) Lwr−wrC1⋅C2C1+C2=0
From (3) we determine the resonant frequency wr
(4) wr=L⋅C1⋅C2C1+C2
The initial charges of the capacitors do not affect the resonant frequency of the circuit. Expression (4) corresponds to the well-known formula of the oscillatory contour wr=LC1 if we take into account (5) C=C1+C2C1⋅C2 for the capacity of serial capacitors.
This formula can be obtained from the following consideration. When connected in series, electrical voltages of the components are added to each other. U=U1+U2 . According to the capacity definition we have U1=C1Q1, U2=C2Q2 and If they change over time U(t)=C1Q1(t)+C2Q2(t) . The current or what is the same amount of charge flowing per unit of time through the capacitors in a serial circuit is the same i.e. I(t)=dtdQ1(t)=dtdQ2(t) . This means that
(6) U˙(t)=I(t)⋅(C11+C21) . The voltage in the AC circuit is written as U(t)=U(w)⋅eiwt and has a time derivative U˙(t)=iw⋅U(w)⋅eiwt substitute to (6) we get
(7) U(w)=I(w)⋅(iwC11+iwC21) when I(t)=I(w)⋅eiwt , and the multiplier eiwt has decreased.
Using the definition of impedance Z(w)=I(w)U(w) we get (1) and (5).
Answer: The formula for the resonant frequency is wr=L⋅C1⋅C2C1+C2
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