Answer:
a) 4.7 kΩ, +/- 5%
b) 2.0 MΩ, +/- 20%
Explanation:
a) If the resistor has the following combination of color bands:
1) Yellow = 1st digit = 4
2) Violet = 2nd digit = 7
3) Red = multiplier = 10e2
4) Gold = tolerance = +/- 5%
this means that the resistor has 4700 Ω (or 4.7 kΩ), with 5% tolerance.
b) Repeating the process for the following combination of color bands:
1) Red = 1st digit = 2
2) Black = 2nd digit = 0
3) Green = multiplier = 10e5
4) Nothing = tolerance = +/- 20%
This combination represents to a resistor of 2*10⁶ Ω (or 2.0 MΩ), with +/- 20% tolerance.
Answer:
The energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Explanation:
Given;
number of turns, N = 179
radius of the circular coil, r = 3.95 cm = 0.0395 m
resistance, R = 10.1 Ω
time, t = 0.163 s
magnetic field strength, B = 0.573 T
Induced emf is given as;
where;
ΔФ is change in magnetic flux
ΔФ = BA = B x πr²
ΔФ = 0.573 x π(0.0395)² = 0.002809 T.m²
According to ohm's law;
V = IR
I = V / R
I = 3.0848 / 10.1
I = 0.3054 A
Energy = I²Rt
Energy = (0.3054)² x 10.1 x 0.163
Energy = 0.1536 J
Energy = 153.6 mJ
Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Answer:
Explanation:
Obtain the following properties at 6MPa and 600°C from the table "Superheated water".
Obtain the following properties at 10kPa from the table "saturated water"
Calculate the enthalpy at exit of the turbine using the energy balance equation.
Since, the process is isentropic process
Use the isentropic relations:
Calculate the enthalpy at isentropic state 2s.
a.)
Calculate the isentropic turbine efficiency.
b.)
Find the quality of the water at state 2
since at 10KPa << at 10KPa
Therefore, state 2 is in two-phase region.
Calculate the entropy at state 2.
Calculate the rate of entropy production.
since, Q = 0
The process of using magnetic fields to produce voltage.