My teacher doesn't say how much power is being drawn by the device plugged in therefore this question doesn't seem answerable to me.

Does this question make sense to any of you?

Sorry for the stupid question.

 

You can calculate from the resistance of the conductors using the allowed voltage drop.
Max.

 

It makes perfect sense. How much power has to be consumed to give you the 5VRMS drop? Simple E=IR stuff.

 

It makes perfect sense. How much power has to be consumed to give you the 5VRMS drop? Simple E=IR stuff.

But he doesn't give a load resistance so how am I supposed to calculate the total current? Obviously if it was just the wires in the street then I could do 5 / 0.2 to get 25 amps but he's implying there is a third component which are the wires inside the house which are also dropping part of the voltage so now it becomes 5 / (0.2 + R). And surely that requires some other bit of information such as the load.

Perhaps I am misreading the question and he actually does just want me to do 5 / 0.2 but that seems to easy.

 

25 Amps max, I = V/R,, = 5V/0.2 ohms

Max power = 25Amp x 225V =? you work it out...

 

What do you mean by "He doesn't give a load resistance"? It's right there in the problem - 0.2Ω. What current through the 0.2Ω resistor will give you a voltage drop of 5VRMS? Now, what is the power that will draw that current (P=I⋅V)?

Everything is there - just figure it out. 25Amps is the current. What is the power?

 

My teacher doesn't say how much power is being drawn by the device plugged in therefore this question doesn't seem answerable to me.

Does this question make sense to any of you?

Sorry for the stupid question.

Do the first part of the problem. Sketch the circuit including all of the values noted. For the load resistance, just put a resistor and call it's value "Rload".

Doing that will probably give you the hint you need to solve the problem asked without needing to know Rload.

 

Do the first part of the problem. Sketch the circuit including all of the values noted. For the load resistance, just put a resistor and call it's value "Rload".

Doing that will probably give you the hint you need to solve the problem asked without needing to know Rload.

I did that, I'm 100% sure I misinterpreted the question. I understand now, it was just unclear wording to me.