"Transformer is used to change the voltage".

Engineering

1 answer:

LUCKY_DIMON [66]3 years ago

6 0

Hi! transformer is used to change the voltage through electromagnetic induction.

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A 1000-turn coil of wire 1.0 cm in diameter is in a magnetic field that increases from 0.10 T to 0.30 T in 10 ms. The axis of th

ddd [48]

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

4 0

3 years ago

Considering the analogy between electrical circuit and thermal circuit, show your approach to derive an expression for the therm

Simora [160]

Answer:

Thermal resistance for a wall depends on the material, the thickness of the wall and the cross-section area.

Explanation:

Current flow and heat flow are very similar when we are talking about 1-dimensional energy transfer. Attached you can see a picture we can use to describe the heat flow between the ends of the wall. First of all, a temperature difference is required to flow heat from one side to the other, just like voltage is required for current flow. You can also see that $R_{th}$ represents the thermal resistance. The next image explains more about the parameters which define the value of the thermal resistances which are the following:

Wall Thickness. More thickness, more thermal resistance.
Material thermal conductivity (unique value for each material). More conductivity, less thermal resistance.
Cross-section Area. More cross-section area, less thermal resistance.

A expression to define the thermal resistance for the wall is as follows: $R_{th} =\frac{l}{Ak}$ , where l is the distance between the tow sides of the wall, that is to say the wall thickness; A is the cross-section area and k is the material conducitivity.