1. An RLC circuit consists of a resistor, inductor, and a capacitor in series along with a battery as their components. Expressed from total energy, we see the total charge as determined by the differential equation dd²q L +R- + dt2 dt Show that from the general solution for a harmonic oscillator under a damped oscillation with the expression that is analogous, d²x dx +b + kx = 0, dt m- dt² gives us the solution of charge as q(t) = Qmaze¬R/2L cos wat, %3D where the angular frequency of the circuit is 271/2 R 1 wd LC 2L Hint: Start with working through finding a general solution for the harmonic oscillator for mechanical energy. Apply it to Electromagnetic energy to find the solution above. We can find that this type of solution is similar.

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1. An RLC circuit consists of a resistor, inductor, and a capacitor in series along
with a battery as their components. Expressed from total energy, we see the
total charge as determined by the differential equation
dd q
dq
+R
+
= 0
dt2
dt
Show that from the general solution for a harmonic oscillator under a damped
oscillation with the expression that is analogous,
d²x
dx
+ b
dt2
dt
= 0,
m
+ kx =
gives us the solution of charge as
,-R/2L
q(t) = Qmaze
cos wat,
where the angular frequency of the circuit is
211/2
R
1
= PM
LC
2L
Hint: Start with working through finding a general solution for the harmonic
oscillator for mechanical energy. Apply it to Electromagnetic energy to find the
solution above. We can find that this type of solution is similar.
Transcribed Image Text:1. An RLC circuit consists of a resistor, inductor, and a capacitor in series along with a battery as their components. Expressed from total energy, we see the total charge as determined by the differential equation dd q dq +R + = 0 dt2 dt Show that from the general solution for a harmonic oscillator under a damped oscillation with the expression that is analogous, d²x dx + b dt2 dt = 0, m + kx = gives us the solution of charge as ,-R/2L q(t) = Qmaze cos wat, where the angular frequency of the circuit is 211/2 R 1 = PM LC 2L Hint: Start with working through finding a general solution for the harmonic oscillator for mechanical energy. Apply it to Electromagnetic energy to find the solution above. We can find that this type of solution is similar.
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