Question

(1)An LC circuit contains a $20mH$ inductor and a $50\mu F$ capacitor with an initial charge of $10mC$. The resistance of the circuit is negligible. Let the instant the circuit is closed be $t=0$. What is the total energy stored initially? Is it conserved during LC oscillations?

(2) An LC circuit contains a $20mH$ inductor and a $50\mu F$ capacitor with an initial charge of $10mC$. The resistance of the circuit is negligible. Let the instant the circuit is closed be $t=0$. What is the natural frequency of the circuit?

(3)An LC circuit contains a $20mH$ inductor and a $50\mu F$ capacitor with an initial charge of $10mC$. The resistance of the circuit is negligible.Let the instant the circuit is closed be $t=0$. At what time is the energy stored
(i)completely electrical (i.e., stored in the capacitor)?
(ii)completely magnetic (i.e., stored in the inductor)?

(4)An LC circuit contains a $20mH$ inductor and a $50\mu F$ capacitor with an initial charge of $10mC$. The resistance of the circuit is negligible.Let the instant the circuit is closed be $t=0$. At what times is the total energy shared equally between the inductor and the capacitor?

(5)An LC circuit contains a $20mH$ inductor and a $50\mu F$ capacitor with an initial charge of $10mC$. The resistance of the circuit is negligible.Let the instant the circuit is closed be $t=0$.

If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?

Answer 1

(1)Given,

Inductance $L=20mH=20\times {10}^{-3}H$

Capacitance of the capacitor

$C=50\mu F=50\times {10}^{-6}F$

Initial charge on the capacitor,

$Q_i=10mC=10\times {10}^{-3}C$

The total energy stored initially is given by, $E=\frac{q}{2C}$

$E=\frac{Q_i^2}{2C}$

Substituting the values,
$E=\frac{1}{2}\times \frac{(10\times {10}^{-3}{)}^2}{50\times {10}^{-6}}=1J$

Yes, this energy is conserved during LC oscillations.
Final Answer: $1J$, Yes

(2) Given,

Inductance $L=20mH=20\times {10}^{-3}H$

Capacitance of the capacitor

$C=50\mu F=50\times {10}^{-6}F$

Natural frequency of the circuit is given by$Q_i=10mC=10\times {10}^{-3}C$

$f=\frac{1}{2\pi \sqrt{LC}}$

Substituting the values, we get
$f=\frac{1}{2\times 3.14\sqrt{20\times {10}^{-3}\times 50\times {10}^{-6}}}$

$f=159.2Hz$

Natural angular frequency of the circuit is given by

$\omega =2\pi f$

Substituting the values, we get

$\omega =2\times 3.14\times 159.2=1000rad/s$

Final Answer:

$f=159.2Hz$
$\omega =1000rad/s$

(3)Step 1: Completely electrical (i.e., stored in the capacitor)

Let at any instant the energy stored is completely electrical.

The charge on the capacitor

$Q=Q_0\cos \omega t$

$Q=Q_0\cos \frac{2\pi }{T}.t$ ....(i)

$Q$ is maximum when it is equal to $Q_0$, only if

$\cos \frac{2\pi }{T}t=\pm 1=\cos n\pi$
Or
$t=\frac{nT}{2},$ where $n=0,1,2,\mathrm{3.......}$

Then $t=0,\frac{T}{2},T,\frac{3T}{2},2T....$

Thus, the energy stored is completely electrical at $t=0,\frac{T}{2},T,\frac{3T}{2}....$

Step 2: Completely magnetic (i.e., stored in the inductor)

Let at any instant, the energy stored is completely magnetic as when the electrical energy across the capacitor is zero.
$Q=0$

$Q_0\cos \frac{2T}{T}=0$

$\therefore \cos \frac{2T}{T}.t=0$

$t=\frac{n\pi }{2}ort=\frac{n\pi }{4}$, where $n=1,3,\mathrm{5....}$

It happens if $t=\frac{T}{4},\frac{3T}{4},\frac{5T}{4}....$

Thus, the energy stored is completely magnetic at
$t=\frac{T}{4},\frac{3T}{4},\frac{5T}{4}....$

Final Answer: (i)$t=0,\frac{T}{2},T,\frac{3T}{2}....$

(ii)$t=\frac{T}{4},\frac{3T}{4},\frac{5T}{4}....$

(4) Let $Q$ be the charge on the capacitor when total energy is equally shared between the capacitor and the inductor at time $t$.

When total energy is equally shared between the inductor and capacitor,

Energy stored in the capacitor $=\frac{1}{2}$ (maximum energy )

$=\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}\left(\frac{1}{2}\frac{Q_0^2}{C}\right)$

$\Rightarrow Q=\frac{Q_0}{\sqrt{2}}$
Using the equation,

$Q=Q_0\cos \frac{2\pi }{T}t$

$\Rightarrow \cos \frac{2\pi }{T}t=\frac{1}{\sqrt{2}}$

$\Rightarrow \cos \frac{2\pi }{T}t=\cos (2n+1)\frac{\pi }{4}$

$\Rightarrow t=(2n+1)\frac{T}{8}$

$\Rightarrow t=\frac{T}{8},\frac{3T}{8},\frac{5T}{8}....$ etc

Final Answer : $t=\frac{T}{8},\frac{3T}{8},\frac{5T}{8}....$

(5) Given,

Inductance $L=20mH=20\times {10}^{-3}H$

Capacitance of the capacitor

$C=50\mu F=50\times {10}^{-6}F$

Initial charge on the capacitor,

$Q_i=10mC=10\times {10}^{-3}C$

The total energy stored initially is given by, $E=\frac{q^2}{2C}$

$E=\frac{Q_i^2}{2C}$

Substituting the values,
$E=\frac{1}{2}\times \frac{(10\times {10}^{-3}{)}^2}{50\times {10}^{-6}}=1J$

$R$ damps out the LC oscillations eventually. So if a resistor is inserted in the circuit, then total energy is eventually dissipated as heat.

Thus, the energy dissipated as heat is $1J$
Final Answer: $1J$