TrueTears question thread

[TrueTears]

ii) and iii) what do you guys get?

[Fireworks]

6, 5, 5?

[TrueTears]

Why not 5.5 for ii and iii?

Me and /0 both got 5.5

The current flows through only 1 wire goes to the transformer then returns to the power supply through the other wire.

Thus the power loss

R should only be 2 and not 4.

[/0]

Yeah if it was asking "what is the voltage delivered at the input to the primary coil of the transformer" then that would just be

right? not

[TrueTears]

Yeah if it was asking "what is the voltage delivered at the input to the primary coil of the transformer" then that would just be

right? not

Exactly that's what I thought as well...

Let's ask BW tomorrow?

[Fireworks]

I also don't get the idea behind that but what I do know is
when I do these calculations, you must include resistance of
both of the wires, therefore a total of 4 ohms.

Voltage drop across the line = I x R = 0.5 x 4 = 2vdrop.

Therefore at V1, P=VI = 10*0.5 = 5w


Sorry, I only got a B+ on mid year so I'm not all that
great at physics but that's my understanding..

[TrueTears]

Thanks heaps for the help yeah that's what originally thought, but then as I thought deeper I began to wonder why you'd use both resistance haha

[dejan91]

You can't just neglect a wire like that, can you?

EDIT: yeah haha after looking into it deeper, I get what you mean, because voltage is being measured from the one wire, so resistance is only 2.

[TrueTears]

You can't just neglect a wire like that, can you?

I know but it doesn't make sense to add the resistance of both. The current can travel whereever from the left terminal to the right or right to the left, but it only passes through ONE wire then the next step in its journey is the transformer then it returns to the other terminal through the OTHER wire. Thus at the transformer it should only have dropped some power over the resistance of ONE wire and not both.

[/0]

You can think of it like this:

________after wire resist______stepped down________after globe_______step up_____after wire resistance
12V---------11V------------------2.2V-----------------0.2V---------------1V----------------0V

If you're looking for the input to the primary coil of the transformer I think it would be 11V

[TrueTears]

You can think of it like this:

________after wire resist______stepped down________after globe_______step up_____after wire resistance
12V---------11V------------------2.2V-----------------0.2V---------------1V----------------0V

If you're looking for the input to the primary coil of the transformer I think it would be 11V

Yeah I had 11 V as well, but the exam says 10 V rofl, let's check tomoz.

[dejan91]

You can't just neglect a wire like that, can you?

I know but it doesn't make sense to add the resistance of both. The current can travel whereever from the left terminal to the right or right to the left, but it only passes through ONE wire then the next step in its journey is the transformer then it returns to the other terminal through the OTHER wire. Thus at the transformer it should only have dropped some power over the resistance of ONE wire and not both.

Exactly what I just thought. I think what VCAA are trying to say is that because current is traveling in both directions, it experiences resistance and thus power loss in both wires, and then the total voltage loss is the sum of these. It doesn't really make sense now when you think about it. It's like a parallel circuit - voltage is distributed evenly...

[almostatrap]

the circuit has to 'know' not to use all the power/voltage so the current can make it home. ahah electricity is crazy

[TrueTears]

the circuit has to 'know' not to use all the power/voltage so the current can make it home. ahah electricity is crazy

Yeah ok that works well for the voltage, what about power then? It didn't use up all the power after passing through 1 wire heh

[littlecherry25]

Hey...
can someone explain to me Question 12 of the VCAA 2008 paper
the question about the flood light?
the assesor's report sayd "the potential difference across the floodlight was in the same ratio as the resistance = (3/4) x 12 = 9"
Huh? where did the 3/4 come from ?
man i've forgotten basic circuit theory stuff, didn't know it was relevant to unit 4 :S