Challenge

Project - Analysis of a practical automotive wiring circuit

Ans : 1 )

% The major elements given in diagram are as follows :

1 ) Generator;

2 ) Battery;

3 ) Starter;

4 ) Distributor;

5 ) Igition switch;

6 ) Foot selector switch;

7 ) Ignition coil;

8 ) Breaker.

Ans : 2 )

1) Battery

Battery is called as heart of any Battery operated system.
Battery is a device which converts chemical energy of cells into electrical energy.
In an automobile a battery is used to supply current to all the electrical equipments.
The battery houses no. of cells which stores the energy when we charge the battery and then we can use it as per our need.
There are many different types of batteries such as lead-acid batteries, lithium-ion batteries, nickel-cadmium bateries and many more.

2 ) Generator

Generator is a device which converts mechanical energy into electrical energy.
It hepls to charge battery and to provide electrical current to different electrical and electronics components when engine is running.
It takes the input energy from engine in from the rotating shaft.
Which in turn rotates the generator shaft and produces current.

3 ) Iginition coil

An ignition coil is used in the ignition system of a engine to increase the battery voltage to the much higher voltage required to make the spark plug work.
The spark plugs then use this electricity to inginite the air fuel mixture coming from carburetor.
The ignition coil itself work as step up transformer to increase the low voltage to high voltage.
Ignition coil consists of primary winding, secondary winding and iron core.
Low voltage current is given to primary winding and it gets transformed in the ignition coil to high voltage, which is received from secondary winding.
Then this high voltage current is given to spark plugs directly or distribution system which distributes the current or gives the signal to the perticular spark plug as per requirement.

Ans : 3 )

In order to select fuse for the load of 10A we should select the fuse whose rating should be slightly more than that of the load or the current which will system will draw.
Thus, for total load of 10A from battery, the fuse rating should not be less than 10 A.

Ans : 4 )

To power this load for a reasonable amount of time, you would need a battery with a capacity of at least 100Ah (41.67A x 2.5 hours).
To determine the voltage and capacity of the battery required for a load of 500W, we should first find the current.
For this assuming a 12V battery, the current drawn could be calculated with the formula:

I = W/V

= 500/12

I = 41.67 A

Ans : 5 )

Ignition swith and light swith are connected in parallel.
Because we know parallel connection is deployed to allow multiple connections to the components with same power sourse while maintaining the same voltage across each component.
By connecting them in parallel we can ensure that each componenet receives correct voltage while allowing each component to draw current it needs

Ans : 6 )

In the above circuit diagram we can observe that two tail lamps are connected in series.
The reason behind series connection is that the current requirement of both the tail lamps is same and they can work with the provided potential differece.

Ship work like this
in your own course.