a) An LC circuit is shown in the figure above. The 44 pF capacitor is initially

Question

a) An LC circuit is shown in the figure above.
The 44 pF capacitor is initially charged by the
6 V battery when S is at position a. Then S
is thrown to position b so that the capacitor
is shorted across the 16 mH inductor.

What is the maximum value for the oscillating

current assuming no resistance in the
circuit?
Answer in units of A.

 

b) What is the natural angular frequency of the
circuit?
Answer in units of rad/s.

 

c)What is the maximum energy stored in the
magnetic field of the inductor?
Answer in units of J.

Explanation / Answer

The frequency                      f = 1/2LC                        = 1/2(16*10^-3H)(44*10^-12)                        = 1.897*10^5 Hz When the switch is thrown to battery the charge on the capacitor         Q = CV = (44*10^-12)(6V) = 2.64*10^-10C The maximum current I = Q = 2f Q                                                   = 2(1.897*10^5Hz)(2.64*10^-10C)                                                  = 3.15*10^-4 A (b) The natural frequency                              f = 1.897*10^5 Hz angular frequency = 2(1.897*10^5Hz                                     = 1.191*10^6 rad/s (c) The energy stored in the inductor              U = 1/2 Li^2                  = (0.5)(16*10^-3 )(3.15*10^-4)^2                  = 7.94*10^-10 J         

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---