Saved Bookmarks
| 1. |
An LC circuit contains a 20 mH inductor and a 50 muFcapacitor with an initial charge of 10 mС. 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. If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat? |
|
Answer» Solution :a. Total initial energy ` = (Q^2)/(2C) = ((10 xx 10^(-3))^2)/(2 xx 50 xx 10^(-6) )= 1J` Yes . If R = 0 , then sum of energies for L and C is conserved b. `omega = (1)/(sqrtLC) = (1)/(sqrt(20 xx 10^(-3) xx 50 xx 10^(-3) )) = 1000 rad//s ` ` v = (omega)/(2pi) = (1000)/(2 xx 3.14) = 160 Hz` c. `q=q_0 cos omega t and u = (q^2)/(2C) `. then i.Energy stored is completely electrical at t = 0 , `T/2, T, (3T)/(2) ` ,......... ii. Electrical energy is zero and energy storedis purely magnetic energy at `t = T/4 , (3T)/(4), (5T)/(4)` ,......... ` T = 1/f = (1)/(160) = 6.2 ms ` d. Energy is shared equally between L and C at `T/8 , (3T)/(8), (5T)/(8) ,...............` SINCE`E = (q_0^2 cos^2 omega t)/(2C)` if `omega t = 45^@, cos omega t= (1)/(sqrt2)` ` THEREFORE E = 1/2 . (q_0^2)/(2C) = 1/2 ` of total energy . e.Total intial energy 1J will be lost as heat due to joule heating effect in the resistor . |
|