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We had to change to 200 and redo the process as we did in the past tries. step by step.After finishing we have to calculate the power P=I.V dissipated in the resistor for each measurement and male a plot of power P vs RL. From this plot eyeball the load resistance, which maximizes the power output. DATA:Pl = (lo/ Ro Rl MY)^2 . Rl mx
Pl= Vo^2/ (Ro + Rmx) = Pl max / p gen . 100100 = Pl mx = (10v/117.9 + 100)^2 .120= 0.248 200= Pl mx =(10v/400)^2 .200= 0.125100= P gen =100/100+120=0.455200=P gen =100/200+200=0.25Error percent: (0.248/0.455) *100% = 54.5% Results:Ro = 100 ohms V=x (A/(B+X)) ^ 2 A: 10.11 +/- 0.04021 B: 117.9 +/- 0.8966RMSE: 0.002258R0= 200 ohms V=x(A/(B+X)) ^2A: -10.05 +/- 0.05228B: 201.9 +/- 1.869 RMSE: 0.001688Discussion:To begin, In this experiment we used a DC power supply to examine the
load of resistance at which power is optimized. In order to do this Resistance was changed (with a constant internal resistance) and voltage V and current I were measured in these experiments. The measurements were taken through a number of calculations and generated two graphs power P versus RLwere created.Graph 1 shows the data acquired when the internal resistance was set to 100Ω and the load resistance was increased from 0-1000Ω and the voltage was measured in volts. Graph 2 shows the same data but for the internal resistance set at 200Ω. These two graphs helped show the relationship between load resistance and power. The relationship between the two helped prove that the