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
The heat of the fire is warming her skin by making the air hot while drinking the hot chocolate warms her from inside of her
The refractive index of a material is a dimensionless number that describes how fast light travels through the material. It is defined as n={\frac {c}{v}}, where c is the speed of light in vacuum and v is the phase velocity of light in the medium.
the ratio of the velocity of light in a vacuum to its velocity in a specified medium.
(b) Always act on the different bodies in opposite directions
