A V = 37.0 V power supply fully charges a capacitor with capacitance C = 1.03 µF. The capacitor is then connected to a L = 86.7 mH inductor. Find the maximum current (in A) in the resulting oscillations.
Question
A V = 37.0 V power supply fully charges a capacitor with capacitance C = 1.03 µF. The capacitor is then connected to a L = 86.7 mH inductor. Find the maximum current (in A) in the resulting oscillations.
Solution
The energy stored in a capacitor is given by the formula:
E = 1/2 * C * V^2
where: E is the energy, C is the capacitance, and V is the voltage.
First, we calculate the energy stored in the capacitor:
E = 1/2 * 1.03 * 10^-6 F * (37.0 V)^2 = 0.00071105 Joules
The energy stored in the capacitor will be completely transferred to the inductor in the form of magnetic energy at the peak of the current. The energy stored in an inductor is given by the formula:
E = 1/2 * L * I^2
where: E is the energy, L is the inductance, and I is the current.
We can set the two energy equations equal to each other and solve for I:
0.00071105 J = 1/2 * 86.7 * 10^-3 H * I^2
Solving for I gives:
I = sqrt((2 * 0.00071105 J) / (86.7 * 10^-3 H)) = 0.036 A
So, the maximum current in the resulting oscillations is 0.036 A.
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