Question

A9: Current through a Resistor Lab Goal: Construct a graph of Electric Current vs. Electric Potential Difference Must Haves: 1. Sketch of the experimental setup illustrating how circuit was actually setup, including how measurement devices were connected. Identify name of each circuit element and measurement device. II. Annotated electric potential vs path graph: Label the high and low potential side, the direction of current flow, and any changes, in electric potential. Identify key junctures around the circuit with letters such as A, B, C, D Tee 0. II. Neatly organized data table: Electric Potential Difference and Current IV. Graph and Linear Fit Equation: Current vs Electric Potential Difference

Answer 1

0.05 - data points from table linear fitting 0.04 0.03 | (Amp) y = a + b*x No Weighting 0.02 Equation Weight Residual Sum of Squares Adj. R-Square 6.99654E-35 1 Value Standard Error 0.01 -3.46945E-18 3.25433E-18 | (Amp) Intercept Slope 0.01 1.07243E-18 0.00 0 1 2 1 3 4 5 V (Volts)

VA А www VB B R E L LA V AD E-> R + Resistor Voltmeter (measures Voltage). Ammeter (measures current) de battery (Power supply) Voltmeter is is connected in parallel with the resistor "R". Ammeter is connected is series with the resistor. Potential at the point A iS VA Potential at the point B is VB omd VAT VB

so according to ohm's Law, (VA-VB) the potential difference generates a Current I through the resistor R. Current I is directly po proportional to the voltage difference (VA-VR). I & [VA-VB) VA-VB 1 = R (VA-VA) = IR Say, VA VB = v V=IR

V - IR 1 = & (V) Consider, 1 न Variable V as X Variable (slope) and 호 R y = moe This is equation of a straight line. The slope slope of the straight line is 호 m = 11 From fitting, slope 0.01 0.01 = R ED R = loo Ω.