Respuesta :
When a charged capacitor is coupled to an inductor, the electrical current and charge in the circuit experience electrical LC oscillations. The capacitor's initial charge, known as qm, is the electrical energy it stores. It is denoted by the equation U E = 1 2 q m 2 C.
Calculation:
(a) From V=IX C
we find ω=I/CV.
The period is then, T=2π/ω=2πCV/I=46.1μs.
(b) The capacitor can store a maximum amount of energy of
U E = 21
CV 2 = 21 (2.20∗10 −7 F)(0.250V) 2 =6.88∗10 −9 J.
(c) The inductor's maximum energy storage capacity is also U.
B =LI 2 /2=6.88nJ.
(d) We apply V=L(di/dt) max
We can substitute L=CV 2 /I 2
combining what we found in part (a)) into equation (as written above) and solve for (di/dt)
max
Our result is,( dtdi )
max = LV = CV 2 /I 2V
= CVI 2
= (2.20∗10 −7 F)(0.250V)(7.50∗10 −3 A) 2
=1.02∗10 3 A/s.
(e) The derivative of UB= 21 Li 2 leads to,
dU B/dt =LI 2 ωsinωtcosωt= 21 LI 2 ωsin2ωt,
Therefore,
dU B / dt
max = 21 LI 2 ω= 21
IV= 21 (7.50∗10 −3 A)(0.250V)=0.938mW.
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