7.12 An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0.

(a) What is the total energy stored initially? Is it conserved during LC oscillations?

(b) What is the natural frequency of the circuit?

(c) 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)?

(d) At what times is the total energy shared equally between the inductor and the capacitor?

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

5 Views | Posted 4 months ago

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1 Answer

Answered by

Payal Gupta | Contributor-Level 10

4 months ago

7.12 Inductance of the Inductor, L = 20 mH = 20 $\times$ $10^{- 3}$ H

Capacitance of the capacitor, C = 50 $μ F$ = 50 $\times 10^{- 6}$ F

Initial charge of the capacitor, Q = 10 mC = 10 $\times 10^{- 3}$ C

The total energy stored initially at the circuit is given as

E = $\frac{1}{2}$ $\times \frac{Q^{2}}{C}$ = $\frac{1}{2}$ $\times \frac{{(10 \times 10^{- 3})}^{2}}{50 \times 10^{- 6}}$ = 1 J

Hence, the total energy stored in the LC circuit will be conserved because there is no resistor connected in the circuit.

Natural frequency of the circuit is given by the relation

$ν = \frac{1}{2 π \sqrt[]{L C}}$ = $\frac{1}{2 π \sqrt[]{20 \times 10^{- 3} \times 50 \times 10^{- 6}}}$ = 159.15 Hz

Natural angular frequency, $ω = \frac{1}{\sqrt[]{L C}}$ = $\frac{1}{\sqrt[]{20 \times 10^{- 3} \times 50 \times 10^{- 6}}}$ = 1000 rad/s

(i) Total time period, T = $\frac{1}{ν}$ = $\frac{1}{159.15}$ = 6.28 $\times 10^{- 3}$ s = 6.2

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7.12 Inductance of the Inductor, L = 20 mH = 20 $\times$ $10^{- 3}$ H

Capacitance of the capacitor, C = 50 $μ F$ = 50 $\times 10^{- 6}$ F

Initial charge of the capacitor, Q = 10 mC = 10 $\times 10^{- 3}$ C

The total energy stored initially at the circuit is given as

E = $\frac{1}{2}$ $\times \frac{Q^{2}}{C}$ = $\frac{1}{2}$ $\times \frac{{(10 \times 10^{- 3})}^{2}}{50 \times 10^{- 6}}$ = 1 J

Hence, the total energy stored in the LC circuit will be conserved because there is no resistor connected in the circuit.

Natural frequency of the circuit is given by the relation

$ν = \frac{1}{2 π \sqrt[]{L C}}$ = $\frac{1}{2 π \sqrt[]{20 \times 10^{- 3} \times 50 \times 10^{- 6}}}$ = 159.15 Hz

Natural angular frequency, $ω = \frac{1}{\sqrt[]{L C}}$ = $\frac{1}{\sqrt[]{20 \times 10^{- 3} \times 50 \times 10^{- 6}}}$ = 1000 rad/s

(i) Total time period, T = $\frac{1}{ν}$ = $\frac{1}{159.15}$ = 6.28 $\times 10^{- 3}$ s = 6.28 ms

Total charge on the capacitor at time t is given by

Q’ = Q $\cos ? \frac{2 π}{T}$ t

For energy stored in electrical, we can write Q = Q’

Hence, it can be inferred that the energy stored in the capacitor is completely electrical at time, t = 0, $\frac{T}{2}$ , T , $\frac{3 T}{2}$ , ……

(ii) Magnetic energy is the maximum when electrical energy Q’ s equal to zero.

Hence it can be inferred that the energy stored in the capacitor is completely magnetic at time, t = $\frac{T}{4}, \frac{3 T}{4}, \frac{5 T}{4}$ …….

Let Q’ be the charge on the capacitor when total energy is equally shared between the capacitor and the inductor a time t. When total energy is equally shared between the inductor and capacitor, the energy stored in the capacitor = $\frac{1}{2}$ (maximum energy)

= $\frac{1}{2} \frac{{Q'}^{2}}{C}$ = $\frac{1}{2}$ ( $\frac{1}{2} \frac{Q^{2}}{C}$ ) = $\frac{1}{4} \frac{Q^{2}}{C}$