Question

Lc circuit consists of a 20.0 mh inductor and a 0.400 µf capacitor. if the maximum instantaneous current is 0.500 a, what is the greatest potential difference across the capacitor

Answer 1

The governing equation for this situation is the energies in both the inductor and the capacitor. These two energies must be equal and are shown by the equation below:

LI^2/2 = CV^2/2,
Where
L = 20.0 mH = 20*10^-3 H
I = 0.5 Amps
C=0.4 uf = 0.4*10^-6 F

Substituting;
(20*10^-3*0.5^2)/2= (0.4*10^-6*V^2)/2
0.0025 = 2*01^-7V^2
V^2 = 0.0025/(2*10^-7) = 12500
V = Sqrt (12500) = 111.8 V

The greatest potential across the capacitor is 111.8 V.