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gulaghasi [49]
3 years ago
11

When the ignition is first turned on a click noise is heard from under the hood of a vehicle equipped with ETC. Technician A say

s that this is the normal operation of the ETC​ self-test. Technician B says that the TAC motor should not move unless the engine is running. Which technician is​ correct?
Physics
1 answer:
Ierofanga [76]3 years ago
4 0

Answer:

Technician A says that this is the normal operation of the ETC self -test is the correct answer.

Explanation:

An engine control unit (ECU), also widely referred to as an engine control module (ECM), is a type of electronic control device that controls an internal combustion engine with a series of actuators to ensure maximum engine performance.

It achieves so by reading values from a multitude of sensors within the engine bay, translating data using multidimensional feedback maps (the so-called lookup tables) and modifying the actuators.

Mechanically fixed and dynamically regulated by mechanical and pneumatic means, air-fuel combination, ignition time, and idle speed were before ECUs.

As soon as the system gets battery voltage, after ignition is turned, the efi computer makes a self-test of all the actuators and sensors, included the ETC.

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<em>1) The relationship in between the electrical energy carriesd by the transmission wires and the amount of the heat loss in it is due to the reason that when the electricity is flown through the wires there are some resistance found in these wires which creates a disturbance in the efficient flow of electricty.Also we know that current have an heating effect when it is in motion as due to if a large amount or magnitude of electricity is flown through the transmission wires it will carry a larger heat effected and also due to the resistance is provided by the wires and so the process of heat loss takes place.</em>

<em>2)It is important to minimize current in transmission wires due to minimize the heat loss and resistance on flowing electric current to make the system more efficient </em>

<em><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u></em><em> 3)Given Resistance = 250 ohms </em>

<em>Electric potential = 150 volts </em>

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<em>volt^2/Resistance = </em>

<em>=</em><em>(150^2/250)(ohms/volts)</em>

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<em>4)Heat energy (H) = Power(P)×Time(t)</em>

<em>4)Heat energy (H) = Power(P)×Time(t)= (90×2)joules = 180 joul</em><em>e</em><em>s</em>

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