In an oscillating LC circuit, L = 3.40 mH and C = 2.10 µF. At t = 0 the charge o
ID: 1979830 • Letter: I
Question
In an oscillating LC circuit, L = 3.40 mH and C = 2.10 µF. At t = 0 the charge on the capacitor is zero and the current is 2.50 A.(a) What is the maximum charge that will appear on the capacitor? C
(b) At what earliest time t > 0 is the rate at which energy is stored in the capacitor greatest? s
(c) What is that greatest rate? W
Explanation / Answer
similar problem in an oscillating LC circuit L = 3.00 mH and C = 2.70 µF at t = 0 the charge on the capacitor is zero and the current is 2.00 A (a) what is the maximum charge that will appear on the capacitor (b) at what earliest time t > 0 is the rate at which energy is stored in the capacitor greatest and (c) what is that greatest rate (a) given I = 2.00 A C = 2.70 x 10-6 F L = 3.00 mH = 3.00 x 10-3 H the charge at any time is given by q = Q sin?t where Q is the maximum charge on the capacitor and? is the angular frequency so at t = 0, q will be zero then the current willbe i = dq / dt = ? Q cos?t so at t = 0 I = ? Q but we know that ? = 1 / vLC so the maximum charge (Q) will be Q = I / vLC = .............. C (b) the energy stored in the capacitor will be U = q2 / 2 C = Q2 sin2?t /2 C the rate of change is given by dU / dt = Q2 ? sin?t cos?t /C in trigonometry we know that sin2?t = 1 so we get 2?t = p / 2 so t will be t = p / 4? = (p / 4) vLC = ............... s (c) on substituting ? = 2p / T, and sin2?t = 1 in the equation dU / dt = (? Q2 / 2C) sin2?t weget (dU / dt)max = 2 p Q2 / 2 TC =p Q2 / T C so T will be T = 2p vLC = ............ s
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