Answer:
Explanation:
Given that,
A capacitor of capacitance
C = 500pF
Charge on capacitor is
Q = 10μC
Capacitor is then connected to an inductor of inductance 10H
L = 10H
Since we want to calculate the maximum energy stored by the inductor, then, we will assume all the energy from the capacitor is transfer to the inductor
So energy stored in capacitor can be determined by using
U = ½CV²
Then, Q = CV
Therefore V = Q/C
U = ½ C • (Q/C)² = ½ C × Q²/C²
U = ½Q² / C
Then,
U = ½ × (10 × 10^-6)² / (500 × 10^-9)
U = 1 × 10^-4 J
U = 0.1 mJ
So the energy stored in this capacitor is transfers to the inductor.
So, energy stored in the inductor is 0.1mJ
B. Current through the inductor
Energy in the inductor is given as
U = ½Li²
1 × 10^-4 = ½ × 10 × i²
1 × 10^-4 = 5× i²
i² = 1 × 10^-4 / 5
i² = 2 × 10^-5
I = √(2×10^-5)
I = 4.47 × 10^-3 Amps
Then,
I = 4.47 mA
Answer:
i have no idea
Explanation:
it could be B. but i dont belive in myself sooooooooooo.....ehhhh yeah
Elements; 1, 9, 12
Compounds: 3, 7, 11
Mixtures: 2, 4, 5, 6, 8, 10
