At a certain instant, the charge stored by a capacitor in an lc circuit is 60 c and the energy stored by the capacitor is nine sixteenths of its maximum value. if the circuit's frequency of oscillation is (40 / ) hz, what is the maximum current in the circuit?

Complete Question: At a certain instant, the charge stored by a capacitor in an LC circuit is 60 microC and the energy stored by the capacitor is nine sixteenths of its maximum value. If the circuit's frequency of oscillation is (40 /pi) Hz, what is the maximum current in the circuit?

Imax = 6.4 mA

Explanation:

In a oscillating LC circuit, energy is exchanged all time between the capacitor and the inductor, in such a way that any time, the total energy in the circuit is equal to the sum of  the energies stored in both circuit elements.

When the capacitor is fully discharged, the energy is stored only at the inductor, so the current is maximum.

We can write the following equation in this case:

1/2 L*Imax² = 1/2 Li² + 1/2q₀²/C

We have as givens, q₀= 60μC and  f=40/π Hz.

We also know, that when the capacitor is charged with this q₀, the energy stored in it (actually in the electric  field between plates) is 9/16 of its maximum value, which must be equal to the maximum energy stored in the inductor (due to the energy conservation), so, we can write the following equation:

1/2 q₀²/C = 9/16*1/2*L*Imax²

In a pure oscillating circuit, there is a fixed relationship between f, L and C, as follows:

ω₀² = 1/LC ⇒ LC = 1/ω₀²

Simplifying common terms, and manipulating terms, we have:

I²max =16*q₀²*(2*π*f₀)² = 16* (60)²*4*π²*((40)/π)² / 9

⇒ I²max = 16*(60)²*(40)²*4*10⁻¹²/9 A

⇒ I max = ((8*60*40) / 3) * 10⁻⁶ A = 6.4*10⁻³ A = 6.4 mA

when the velocity of a mass on a spring is zero, the acceleration is at a maximum .

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