I have been given the bleow circuit diagram and am stuggling to get my head around what happens with it.

I have been asked what happens when the switch is set to A and also what happens when the switch is set to B.

I have been trying to get my head around it for the last few days and am just confusing myself more and more. I would greatly appreciate it if someone could help me out

 

Show as here your version of what is happens in this circuit.

But first tell me what is the name for this strange component.
Maybe a diode freewheeling diode? If so, you made the error in the diagram.
Diode cathode should be connected to the fuse and anode to collector of a BJT.

 

My fault, it is meant to be a diode and I forgot to draw on the line after the triangle point, it is drawn the correct way around though as the anode is connected to the fuse and cathode to the tranistor and current flows from anode to cathode.

 

it is drawn the correct way around though as the anode is connected to the fuse and cathode to the tranistor and current flows from anode to cathode.

Are you sure about that??

When the switch is in A position nothing happen in the circuit.
The BJT is in cutoff state, because there is no path (from V+) for the base current to open the BJT.

But when you change the position of a switch to point B the interesting things start to happen.

The base see the path for the V+ through:

V+--> R2 --> switch --> base-emiter ---> gnd (V-)

So BJT for sure will be "ON" this mean that collector current will start (want) to flow.

The collector current will flow from:
V+----> relay ---> collector - emitter ----> gnd (V-)
And relay switch LR1 is close and the light bulb lights.

But as you said you are sure that the diode is connect in "correct way".
So when you change the position of a switch to position B the BJT will start to conduct and burn the diode and perhaps maybe the BJT will be burn too. Because we made a beautiful short circuit (V+ ---> Diode----> collector-emitter--->gnd).
End of the story.

 

Im 99.9% sure that i have drawn it the way it was in my test but could you please indulge me in what would happen if the diode was the other way around?

 

The diode is in backwards.

 

Show as here your version of what is happens in this circuit.

But first tell me what is the name for this strange component.
Maybe a diode freewheeling diode?

The diode is there to clamp the voltage kick from the inductance of the relay winding when the relay turns off.

 

The diode will act as a freewheel diode only wen we connect the diode in this way

This diode clamping the voltage induce by the coil when the relay current suddenly drops to zero. (switch change his position form B to A).

In you diagram the diode is connect in wrong way, but we also have a fuse in the diagram. So when we "set" the switch in B position, short-circuit current will start to flow and there is a chance that the fuse will blow-up first before diode/BJT.

 

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This diode clamping the voltage induce by the coil when the relay current suddenly drops to zero. (switch change his position form B to A).

In you diagram the diode is connect in wrong way, but we also have a fuse in the diagram. So when we "set" the switch in B position, short-circuit current will start to flow and there is a chance that the fuse will blow-up first before diode/BJT.

The relay current can't suddenly drop to zero. That's why you need the diode. It provides a path for the inductive relay current (eliminating the large voltage spike otherwise generated) until the inductive energy is dissipated.

I don't think the fuse will blow first. The general rule is "the semiconductor is always faster then the fuse".