Given Information:
Resistance = R = 14 Ω
Inductance = L = 2.3 H
voltage = V = 100 V
time = t = 0.13 s
Required Information:
(a) energy is being stored in the magnetic field
(b) thermal energy is appearing in the resistance
(c) energy is being delivered by the battery?
Answer:
(a) energy is being stored in the magnetic field ≈ 219 watts
(b) thermal energy is appearing in the resistance ≈ 267 watts
(c) energy is being delivered by the battery ≈ 481 watts
Explanation:
The energy stored in the inductor is given by

The rate at which the energy is being stored in the inductor is given by

The current through the RL circuit is given by

Where τ is the the time constant and is given by


Therefore, eq. 1 becomes

At t = 0.13 seconds

(b) thermal energy is appearing in the resistance
The thermal energy is given by

(c) energy is being delivered by the battery?
The energy delivered by battery is

Pretty much any element(in your case sodium) contain these properties.
Atoms can be an ion, but not all ions are atoms. The difference between an atom and an ion has to do with net electrical charge. An ion is a particle or collection of particles with a net positive or negative charge. ... A stable atom contains the same number of electrons as protons and no net charge
<span>A portion of the atmosphere that becomes warmer than surrounding air will expand and rise. The warmer atmosphere the more space between the molecules. Therefore, warmer atmosphere </span><span>expands to allow more space for the molecules. Cool air on the other hand, contracts because the molecules in cool air need less space.</span>
Answer:
a)η = 69.18 %
b)W= 1210 J
c)P=3967.21 W
Explanation:
Given that
Q₁ = 1749 J
Q₂ = 539 J
From first law of thermodynamics
Q₁ = Q₂ +W
W=Work out put
Q₂=Heat rejected to the cold reservoir
Q₁ =heat absorb by hot reservoir
W= Q₁- Q₂
W= 1210 J
The efficiency given as



η = 69.18 %
We know that rate of work done is known as power


P=3967.21 W
Answer:
C. Both A and B
Explanation:
Fuses are rated by the amperage they can carry before heat melts the element. The fuse is ideal for protection against short circuits. Short circuits produce enough amperage to vaporize a fuse element and break connection in one cycle of a 60-cycle system.
Specifically, the voltage rating determines the ability of the fuse to suppress the internal arcing that occurs after a fuse link melts and an arc is produced.