Answer:
Magnetic energy stored in the inductor when all of the energy in the circuit is in the inductor = 0.049 mJ
If all the energy is then transferred into the capacitor, the voltage drop across the capacitor = 0.00572 V = 0.01 V (expressed to the hundredths value)
Explanation:
In an RLC circuit with maximum current of 7mA = 0.007 A
When all of the energy is stored in the inductor, maximum current will flow through it,
Hence E = (1/2) LI²
L = inductance of the inductor = 2 H
E = (1/2) (2)(0.007²) = 0.000049 J = 0.049 mJ
When all the energy in the circuit is in the capacitor, this energy will be equal to the energy calculated above.
And for a capacitor, energy is given as
E = (1/2) CV²
E = 0.000049 J, C = 3 F, V = ?
0.000049 = (1/2)(3)(V²)
V = 0.00572 V = 0.01 V
