How Does Wire Length Affect Voltage in an Electrical Circuit?

AI Thread Summary
Wire length affects voltage in an electrical circuit, as observed in an investigation where voltage readings increased slightly (about 0.1 V) with longer wire lengths (10-100 cm). This phenomenon is attributed to internal resistance and energy loss due to higher resistance in the circuit. If the power supply has internal resistance, it impacts voltmeter readings, especially as current changes. The relationship between voltage and resistance becomes crucial when considering a power pack with finite source resistance. Understanding these dynamics is essential for accurate circuit analysis.
polekitten
Messages
5
Reaction score
0

Homework Statement


We conducted an investigation into wire length, current and resistance but kept track of the voltage rather than relying on the power pack setting. We noticed the voltmeter (measuring across the power pack) reading increased as the length of the wire increased though by only about 0.1 V from 10-100 cm lengths.

Homework Equations

The Attempt at a Solution


Is this something to do with internal resistance and energy being lost due to the higher resistance?
 
Physics news on Phys.org
polekitten said:

Homework Statement


We conducted an investigation into wire length, current and resistance but kept track of the voltage rather than relying on the power pack setting. We noticed the voltmeter (measuring across the power pack) reading increased as the length of the wire increased though by only about 0.1 V from 10-100 cm lengths.

Homework Equations

The Attempt at a Solution


Is this something to do with internal resistance and energy being lost due to the higher resistance?
You have suggested internal resistance. What would happen to our voltmeter readings during the experiment if the power supply had a small internal resistance, such as 1 Ohm? Say the current changes from 0.1 amp to 1 amp as the wire is shortened.
 
You have not defined the power pack electrically. The easiest assumption would be that is is a source of constant current, in which case the relation between V and R should be obvious.

If the power pack has finite (as opposed to infinite) source resistance then that resistance has to be included in your computations of R, given V.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top