Answer to Question #212617 in Electricity and Magnetism for Joramichi

Question

An LC circuit is composed of a 100.0 mH inductor and a 4.0 µF capacitor. If the capacitor can

have a maximum charge of 3.0 µC on its plates,

(a) what is the total energy in the circuit in Joules (J)?

(b) what is the angular frequency of oscillations in the circuit in rads/sec?

(c) what is the maximum current in the circuit in Amperes (A)?

Answer 1

a)

When the charge in the capacitor will be maximum, then the energy in the inductor will be zero

Hence the total energy in the circuit will be $=\frac{Q^2}{2C}$

Now, substituting the values, $=\frac{(3\times 10^{-6})^2}{2\times 4\times 10^{-6}} = \frac{9}{8}\times 10^{-6}J$

b)

Angular frequency of the oscillation in the circuit $\omega = \frac{1}{\sqrt{LC}}$

Now, substituting the values, $\omega = \frac{1}{\sqrt{100\times 10^{-3}\times 4\times 10^{-6}} } rad/sec$

$\Rightarrow \omega = \frac{1}{20\times 10^{-4}\sqrt{10}} rad/sec$

$\Rightarrow \omega =\frac{500}{\sqrt{10}}$ rad/sec

c) Now, applying the energy conservation,

$\frac{LI^2}{2}=\frac{9}{8}\times 10^{-6}$

Now, substituting the values,

$I=\sqrt{\frac{9}{4\times 100\times 10^{-3}}\times 10^{-6}}$

$I=\sqrt{2.25\times 10^{-5}}A$