Question

An RLC circuit with a resistor of R\:=\:1 R = 1 k, capacitor ofC\:=\:3 C = 3 F, and inductor ofL\:=\:2 L = 2 H reaches a maximum current through the inductor of 7 mA. When all of the energy stored in the circuit is in the inductor, what is the magnetic energy stored? (Express your answer in micro-Joules) In the same circuit explained above, if all the energy is then transferred into the capacitor, what voltage drop will there be across the capacitor? (Express your answer in Volts to the hundredths place)

Answer 1

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