For your final exam in electronics, you\'re asked to build an LC circuit that os

Question

For your final exam in electronics, you're asked to build an LC circuit that oscillates at 13 kHz . In addition, the maximum current must be 0.10 A and the maximum energy stored in the capacitor must be 1.5×105 J .

Part A

What value of inductance must you use? Express your answer to two significant figures and include the appropriate units. L=? and what units?

Part B

What value of capacitance must you use?Express your answer to two significant figures and include the appropriate units. C=? and what units?

Explanation / Answer

Well an LC oscillator just uses an inductor and a capacitor to oscillate. When a circuit first turns on, a capacitor is ready to accept the incoming current, but the inductor will oppose it. The opposition will cause the capacitor to react by allowing the inductor to "take over" for a short while. When this happens, the inductor is no longer experiencing an opposition, so it "relaxes". The capacitor can then receive the incoming current again. But this incoming current is again opposed by the inductor. This repeating process will cause an oscillation at a frequency of w = 1/sqrt(L*C), where w is omega, or frequency in terms of radians per second. You'll need to divide this value by 2*pi to get it in terms of hertz. Well for starters, we want this circuit to operate at 13 kHz, or 81.68 krad/s, so 1/sqrt(L*C) = 81.68k Inverting and squaring tells you your value of L*C should be L*C = 1.499*10^-10 Hz.

We now have a general value to work with. Next we know the maximum current must be 100 mA and the capacitor can only have 1.5*10^-5 J (15 uJ). This will allow us to solve for both. The way this oscillator circuit behaves is that it basically transfers power back and forth between the inductor and capacitor. As time goes on, this power would slowly dissipate to internal resistances and such, so the maximum value is what is initially in the circuit. Since the most energy that the capacitor can store is 15 uJ, this is also the most energy the inductor can store as well. We also know the maximum current, which is convenient because an inductor's energy is determined as E = 0.5*L*I^2. Since I = 0.10 A, and E = 15 uJ, plugging values in tells you the inductor must be

L = 15uJ/(0.5*(0.10A)^2) = which is 3000 uH. Now we know an inductance so we can also determine the capacitance.

Again, L*C = 1.499*10^-10, and L = 3*10^-3, so C must be

C = L*C/L = (1.499/3)*10^-7 = 4.997 * 10^-8 F = 49.97 nF.

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