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  • For an LCR circuit driven at frequency \omega, the equation reads (i) Multiply the equation by i and simplify where possible. (ii) Interpret each term physically. (iii) Cast the equation in the form of a conservation of energy statement.

For an LCR circuit driven at frequency \omega, the equation reads

L\frac{di}{dt}+Ri+\frac{q}{C}=v_{1}=v_{m}\; \sin \omega t

(i) Multiply the equation by i and simplify where possible.
(ii) Interpret each term physically.
(iii) Cast the equation in the form of a conservation of energy statement.
(iv) Integrate the equation over one cycle to find that the phase difference between v and i must be acute.

Answers (1)

V=V_{m}\sin \omega t

L\frac{di}{dt}+iR+\frac{q}{C}=V_{m}\sin \omega t

Li\frac{di}{dt}+i^{2}R+\frac{q}{C}i=iV_{m}\sin \omega t

Li\frac{di}{dt}+i^{2}R+\frac{q}{C}i=iV

Power loss = i^{2}R

Rate of change of energy stored in Inductor =

Li\frac{di}{dt}=\frac{d}{dt}\left ( \frac{1}{2}Li^{2} \right )

Rate of change of energy stored in Capacitor =

\frac{q}{C}i=\frac{d}{dt}\left ( \frac{q^{2}}{2C} \right )

\int_{0}^{T}\frac{d}{dt}\left ( \frac{1}{2}Li^{2}+\frac{q^{2}}{2C} \right )dt+\int_{0}^{T}i^{2}R dt=\int_{0}^{T}iV \; dt

0+(+ve)=\int_{0}^{T}iV\; dt>0

 

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