Question

LC CIRCUIT: A 20.00-F capacitor is fully charged by a 100.00-V battery, then disconnected from the battery and connected in series with a 0.280-mH inductor at t=0. Thus, the initial current 1-0. See circuit below after the charged capacitor is connected to the inductor. (a) What is the angular oscillation frequency o of the circuns (b) What is the maximum charge on the positive capacitor plate? (c) What is the energy stored in the capacitor at t - 0? (d) Using conservation of energy, find the current I when the charge reaches one-third the maximum charge. (e) Using conservation of energy, find the maximum current Imax 000000

Answer 1

Answer:

(a) $w=13.363\ rad/sec$

(b) $Q_m=2000\ coul$

(c) $E_{max}=100,000\ J$

(d) $I=25,197.63\ A$

(e) $I_{max}=26,726.12\ A$

Explanation:

LC Circuit

The dynamics of an LC circuit is explained as the energy stored in the capacitor C in the form of an electrical field is transferred to the inductor L as a magnetic field. The energy stored in a capacitor is

$\displaystyle E_c=\frac{CV^2}{2}$

Where V is the potential between its plates where a charge Q is stored. The relation between them is

$\displaystyle C=\frac{Q}{V}$

The energy stored in an inductor of self-inductance L is

$\displaystyle E_c=\frac{LI^2}{2}$

Where I is the current flowing through the inductor. If we apply the principle of conservation of energy, the loss of electric energy in the capacitor will transform into magnetic energy in the inductor and vice-versa.

The angular frequency of oscillation of a LC circuit is

$\displaystyle w=\sqrt{\frac{1}{LC}}$

The question provides the following data

$V_m=100\ V,\ C=20\ F,\ L=0.28\ mH=0.00028\ H$

(a) The angular frequency is

$\displaystyle w=\sqrt{\frac{1}{(0.00028)(20)}}$

$w=13.363\ rad/sec$

(b) When the voltage is at maximum, the charge will also be maximum, its value can be computed solving for Q

$\displaystyle C=\frac{Q_m}{V_m}$

$Q_m=V_mC=100(20)=2000$

$Q_m=2000\ coul$

(c) At t=0, the voltage is at maximum, so the energy stored is

$\displaystyle E_{max}=\frac{CV_m^2}{2}$

$\displaystyle E_{max}=\frac{20\ 100^2}{2}=100,000$

$E_{max}=100,000\ J$

(d) If the charge reaches 1/3 of the initial value, it means 2/3 of the charge were transformed in magnetic energy. Let's call $E_t$ to the transferred electric energy to magnetic energy, and $E_1$ to the remaining electric energy in the capacitor. Knowing $Q_1=1/3Q_m$

$\displaystyle Q_1=\frac{2000}{3}=666.67\ coul$

$\displaystyle V_1=\frac{666.67}{20}=33.33\ V$

$\displaystyle E_1=\frac{(20)\ 33.33^2}{2}=11,111.11\ J$

The energy transfered is

$E_t=100,000-11,111.11=88,888.89\ J$

We can compute the current solving for I in:

$\displaystyle E_t=\frac{LI^2}{2}$

$\displaystyle I=\sqrt{\frac{2E_t}{L}}$

Calculating I

$\displaystyle I=\sqrt{\frac{2(88,888.89)}{0.00028}}$

$I=25,197.63\ A$

(e) To find $I_{max}$, we'll assume all the electric energy is transformed to magnetic, so

$\displaystyle I_{max}=\sqrt{\frac{2E_{max}}{L}}$

$\displaystyle I_{max}=\sqrt{\frac{2(100,000)}{0.00028}}$

$I_{max}=26,726.12\ A$