Question #111441
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

https://www.chegg.com/homework-help/questions-and-answers/case-difficult-read-charges-right-capacitor-q2-q2-q37457186

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 ww can be written as.

(1) Z(w)=iwL+1iwC1+1iwC2Z(w)=iwL+\frac {1}{iwC_1}+\frac {1}{iwC_2}

Find the magnitude of impedance (1)

(2) Z(w)=i(LwC1+C2wC1C2)=LwC1+C2wC1C2|Z(w)|=|i(Lw-\frac{C_1 + C_2}{w C_1\cdot C_2})|=|Lw-\frac{C_1 + C_2}{w C_1\cdot C_2}| .

Resonance in a sequential circuit is achieved at a frequency at which Z(w)min|Z(w)|\to min. In our case of the absence of resistive elements the exact condition must be met Z(w)=0|Z(w)|=0, that is

(3) LwrC1+C2wrC1C2=0Lw_r-\frac{C_1 + C_2}{w_r C_1\cdot C_2}=0

From (3) we determine the resonant frequency wrw_r

(4) wr=C1+C2LC1C2w_r=\sqrt{\frac{C_1+C_2}{L\cdot C_1\cdot C_2}}

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=1LCw_r=\frac{1}{\sqrt{LC}} if we take into account (5) C=C1C2C1+C2C=\frac{C_1\cdot C_2}{C_1+C_2} 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+U2U=U_1+U_2 . According to the capacity definition we have U1=Q1C1U_1=\frac {Q_1}{C_1}, U2=Q2C2U_2=\frac {Q_2}{C_2} and If they change over time U(t)=Q1(t)C1+Q2(t)C2U(t)=\frac{Q_1(t)}{C_1}+\frac{Q_2(t)}{C_2} . 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)=dQ1(t)dt=dQ2(t)dtI(t)=\frac{dQ_1(t)}{dt}=\frac{dQ_2(t)}{dt} . This means that

(6) U˙(t)=I(t)(1C1+1C2)\dot{U}(t)=I(t)\cdot (\frac{1}{C_1}+\frac{1}{C_2}) . The voltage in the AC circuit is written as U(t)=U(w)eiwtU(t)=U(w)\cdot e^{iwt} and has a time derivative U˙(t)=iwU(w)eiwt\dot{U}(t)=iw\cdot U(w)\cdot e^{iwt} substitute to (6) we get

(7) U(w)=I(w)(1iwC1+1iwC2)U(w)=I(w)\cdot (\frac{1}{iwC_1}+\frac{1}{iwC_2}) when I(t)=I(w)eiwtI(t)=I(w)\cdot e^{iwt} , and the multiplier eiwte^{iwt} has decreased.

Using the definition of impedance Z(w)=U(w)I(w)Z(w)=\frac{U(w)}{I(w)} we get (1) and (5).

Answer:  The formula for the resonant frequency is wr=C1+C2LC1C2w_r=\sqrt{\frac{C_1+C_2}{L\cdot C_1\cdot C_2}}


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