In an ideal oscillatory circuit, the inductance of the coil is L = 4mH, the maximum current is I0 = 100mA

| July 9, 2021 | education

In an ideal oscillatory circuit, the inductance of the coil is L = 4mH, the maximum current is I0 = 100mA in it. Find the energy of the electric field of the capacitor at the moment when the current in the coil is I = 50mA.

To find the value of the electric field energy of the capacitor used at a given time, we use the equality: WС = W – WL = L * I0 ^ 2/2 – L * I ^ 2/2 = (I0 ^ 2 – I ^ 2) * L / 2 …

Variables: I0 – maximum current in the coil (I0 = 100 mA = 0.1 A); I is the considered current strength in the coil (I = 50 mA = 0.05 A); L – inductance (L = 4 mH = 0.004 H).

Let’s perform the calculation: WC = (I0 ^ 2 – I ^ 2) * L / 2 = (0.1 ^ 2 – 0.05 ^ 2) * 0.004 / 2 = 15 * 10 ^ -6 J.

Answer: The energy of the electric field of the capacitor used will be equal to 15 μJ.



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