Question

My physics class is working on Ohm's law and we did a lab about series and parallel circuits. While doing the lab report I got stuck on this question below. Please help!

"What contributed to the percentage difference? In other words, account for sources of errors."

This is my data:

Part I

Voltage (V)	Current (A)	Resistance (Ω) R = V/I	% diff.
33	0.33A	100	0
45	0.45A	100	0
60	0.60A	100	0
75	0.75A	100	0
90	0.90A	100	0
120	1.20A	100	0

Part II

Voltage (V)	Current (A)	Resistance (Ω) R = V/I	% diff.
33	0.12A	275	5%
45	0.17A	264.7	5.3%
60	0.22A	272.7	2.7%
75	0.28A	267.8	2.2%
90	0.33A	272.7	2.7%
120	0.44A	272.7	2.7%

Part III:

Voltage (V)	Current (A)	Resistance (Ω) R =V/I	% diff.
33	1.11A	29.7	2.3%
45	1.51A	29.8	2.2%
60	2.02A	29.7	2.3%
75	2.52A	29.7	2.3%
90	3.02A	29.8	2.2%
120	4.03A	29.7	2.3%

Answer 1

Let there be two resistances R1 and R2, connected in either series or parallel combination. Let the resistances be connected by wire 1 and wire 2. Now, the wires connecting these resistances also have some internal resistances, since they have got some length L and thickness d, so that their resistances vary as , Let the two wires connecting resistances R1 and R2 have internal resistances r1 and r2.

Case I : when the resistances are connected in series, the total equivalent resistance is .

Case II : when the resistances are connected in parallel, the wires' resistances (r1 & r2) are also connected in parallel. So, if R_eq be the equivalent resistance, then . Thus even if r1 and r2 is very small in magnitude, they contribute to the total equivalent resistances. So the percentage errors come in the data.