emf generated by the coil is 1.57 V
Explanation:
Given details-
Number of turns of wire- 1000 turns
The diameter of the wire coil- 1 cm
Magnetic field (Initial)= 0.10 T
Magnetic Field (Final)=0.30 T
Time=10 ms
The orientation of the axis of the coil= parallel to the field.
We know that EMF of the coil is mathematically represented as –
E=N(ΔФ/Δt)
Where E= emf generated
ΔФ= change inmagnetic flux
Δt= change in time
N= no of turns*area of the coil
Substituting the values of the above variables
=1000*3.14*0.5*10-4
=.0785
E=0.0785(.2/10*10-3)
=1.57 V
Thus, the emf generated is 1.57 V
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Explanation:
Answer:
a) 23.89 < -25.84 Ω
b) 31.38 < 25.84 A
c) 0.9323 leading
Explanation:
A) Calculate the load Impedance
current on load side = 0.75 p.u
power factor angle = 25.84
= 0.75 < 25.84°
attached below is the remaining part of the solution
<u>B) Find the input current on the primary side in real units </u>
load current in primary = 31.38 < 25.84 A
<u>C) find the input power factor </u>
power factor = 0.9323 leading
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<em>attached below is the detailed solution </em>
Answer:
No
Explanation:
Heat engines are used for converting the heat into mechanical energy which is used for doing mechanical work.
The efficiency of heat engine is the fraction of mechanical energy to the thermal energy. The efficiency can not be 100% as some of the energy always loss due to friction and motion of the body parts of the heat engine.
Explanation:
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