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  • In the LCR circuit shown in Fig 7.7, the ac driving voltage is v=v_{m} sin omega t (i) Write down the equation of motion for q (t). (ii) At t=t_{0}, the voltage source stops and R is short circuited.

In the LCR circuit shown in Fig 7.7, the ac driving voltage is v=v_{m}\sin \omega t

(i) Write down the equation of motion for q (t).
(ii) At t=t_{0}, the voltage source stops and R is short circuited. Now write down how much energy is stored in each of L and C.
(iii) Describe subsequent motion of charges.

 

Answers (1)

a) V=V_{m}\sin \omega t

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

L\frac{d^{2}q}{dt^{2}}+\frac{dq}{dt}R+\frac{q}{C}=V_{m}\sin \omega t

b) Let q=q_{m}\sin \left ( \omega t+\phi \right )

\frac{dq}{dt}=\omega q_{m}\cos (\omega t+\phi )

i_{m}=\frac{V_{m}}{Z}=\frac{V_{m}}{\sqrt{(R)^{2}+(X_{c}-X_{L})^{2}}}

\phi =\tan ^{-1}\left ( \frac{X_{c}-X_{L}}{R} \right )

At t=t_{0}, Resistance is short - circuited and the inductor and capacitor store energy

U_{L}=\frac{1}{2}Li^{2}=\frac{1}{2}L\left [ \frac{V_{m}}{\sqrt{(R)^{2}+(X_{c}-X_{L})^{2}}} \right ]^{2}\sin ^{2} (\omega t_{0}+\phi )

U_{c}=\frac{q^{2}}{2C}=\frac{1}{2C\omega ^{2}}\left [ \frac{V_{m}}{\sqrt{(R)^{2}+(X_{C}-X_{L})^{2}}} \right ]^{2}\; \cos ^{2}(\omega t_{0}+\phi )

(c) When R is short-circuited, the circuit becomes an L-C oscillator. When the capacitor discharges, all its energy goes from capacitor to inductor. Energy oscillates from electrostatic to magnetic and from magnetic to electrostatic.

 

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