Respuesta :
The question is incomplete. The complete question is :
In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?
Solution :
The underdamped RLC circuit
[tex]$v_{t} = ve^{-\frac{R}{2L}t} \cos \omega t$[/tex]
[tex]$\omega = \sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}= \frac{2 \pi}{T}$[/tex]
We know in one time period, v = 2v, at t = T, [tex]$v_t = 3.8 v$[/tex]
so, [tex]$9.8 = 5 e^{-\frac{R}{2L}T} \cos \frac{2 \pi}{T}T$[/tex]
[tex]$e^{-\frac{R}{2L}T} = \frac{3.8}{5} \times 1$[/tex]
[tex]$\frac{R}{2L}T= \ln \frac{5}{3.8}$[/tex]
[tex]$\frac{R}{L}= \frac{2}{1.2 \times 10^{-6}} \ln \frac{5}{3.8}$[/tex]
[tex]$\frac{R}{L} = 457.3 \times 10^3$[/tex]
Now, Q value [tex]$= \frac{1}{R}\sqrt{\frac{L}{C}}$[/tex]
[tex]$=\sqrt{\frac{L}{R^2C}\times \frac{L}{L}}$[/tex]
[tex]$=\sqrt{(\frac{L}{R})^2 \times \frac{1}{LC}}$[/tex]
[tex]$\frac{1}{LC}=27.43 \times 10^{12}$[/tex]
∴ [tex]$Q=\sqrt{\left(\frac{1}{457.3 \times 10^3}\right)^2 \times 27.43 \times 10^{12}}$[/tex]
[tex]$Q=\sqrt{131.166}$[/tex]
= 11.45
