Question

In an oscillating series *RLC* circuit, with a 5.79 ?
resistor and a 17.3 H inductor, find the time required for the
maximum energy present in the capacitor during an oscillation to
fall to half of its initial value.

Answer #1

*The max energy on the capacitor =q _{max}^{2}
/ 2C*

*therefore, the max energy = (1/6) Q ^{2} /2C*

*The maximum charge is given by*

* q _{max}
=Qe^{-Rt/2L}*

* Or
q _{max} / Q = e^{-Rt/2L}*

* Or
ln( q _{max} / Q ) = -Rt / 2L*

*t =( 2L / R) (1/2) ln( 2)*

* t = ln(2) x (L / R) = 0.693 x (17.3 / 5.79) =
2.07 sec*

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E =
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with Em = 45.2 V and
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