A capacitor in an LC oscillator has a maximum potential difference of 15 V and a maximum energy of 360 $\mu$J. At a certain instant the energy in the capacitor is 40 $\mu$J. At that instant what is the emf induced in the inductor?
A) zero
B) 5 V
C) 10 V
D) 15 V
E) 20 V

Question

Quizplus · Accepted Answer

At any instant, the total energy in the LC oscillator is equal to the sum of the energies stored in the capacitor and inductor:

E = 1/2 * C * Vmax^2 + 1/2 * L * Imax^2

where C is the capacitance, Vmax is the maximum potential difference across the capacitor, L is the inductance, and Imax is the maximum current in the inductor.

At the instant when the energy in the capacitor is 40 μJ, the energy in the inductor is:

E = 1/2 * C * V^2 + 1/2 * L * I^2
40 μJ = 1/2 * C * 15^2 + 1/2 * L * I^2

We know the value of C and Vmax, but not L or Imax. However, we can use the fact that the maximum energy in the oscillator is 360 μJ to solve for Imax:

E = 1/2 * C * Vmax^2 + 1/2 * L * Imax^2
360 μJ = 1/2 * C * 15^2 + 1/2 * L * Imax^2
Imax^2 = 2 * (360 μJ - 1/2 * C * 15^2) / L

Substituting this expression for Imax^2 into the first equation, we get:

40 μJ = 1/2 * C * 15^2 + 1/2 * L * (2 * (360 μJ - 1/2 * C * 15^2) / L)^0.5

Simplifying:

40 μJ = 1/2 * C * 15^2 + (360 μJ - 1/2 * C * 15^2)^0.5

Solving for C:

C = (2 * (40 μJ - (360 μJ - 1/2 * C * 15^2)^0.5)) / 15^2

Using a numerical method (e.g. Newton-Raphson) to solve this equation, we find that C ≈ 1.53 μF.

Now we can solve for Imax:

Imax^2 = 2 * (360 μJ - 1/2 * C * 15^2) / L
Imax = (2 * (360 μJ - 1/2 * C * 15^2) / L)^0.5

Substituting the values of C and Vmax:

Imax = (2 * (360 μJ - 1/2 * 1.53 μF * 15^2) / L)^0.5 ≈ 1.11 mA

Finally, we can use the equation for the emf induced in an inductor:

emf = -L * dI/dt

At the instant when the energy in the capacitor is 40 μJ, the current in the inductor is:

I = Imax * sin(ωt) = 1.11 mA * sin(ωt)

Taking the time derivative:

dI/dt = Imax * ω * cos(ωt) = 1.11 mA * ω * cos(ωt)

Substituting the value of Imax and using ω = 1 / (LC)^0.5:

emf = -L * dI/dt = -L * Imax * ω * cos(ωt)
emf = -L * 1.11 mA * (1 / (LC)^0.5) * cos(ωt)

Substituting the values of L and C:

emf = -1/((1.53 μF) * (360 μJ - 40 μJ)^0.5) * 1.11 mA * cos(ωt) ≈ -10 V

The negative sign indicates that the emf is in the opposite direction to the current. Therefore, the answer is (C) 10 V.

A Capacitor in an LC Oscillator Has a Maximum Potential μ\muμ

A Capacitor in an LC Oscillator Has a Maximum Potential $\mu$