Question: 3. Study equation 5. Introduce general variables x and y defined as x=R/r and y=Pyr/E2. Make the substitutions PL = yE2/r and R=xr in equation

 3. Study equation 5. Introduce "general variables" x and y definedas x=R/r and y=Pyr/E2. Make the substitutions PL = yE2/r and R=xrin equation 5. Simplify this expression until you have an equation interms of x and y and no other variables. What is the

3. Study equation 5. Introduce "general variables" x and y defined as x=R/r and y=Pyr/E2. Make the substitutions PL = yE2/r and R=xr in equation 5. Simplify this expression until you have an equation in terms of x and y and no other variables. What is the equation? Sketch the graph of y versus x. what is the peal value of y and the corresponding x value? are these x and y values consistent with your result of question 2?2. Study equation 5 (manual). Assume that the internal resistance is r=11.4W and E=4.0 Volts. Calculate, using a spreadsheet, the P - values for external resistance values R=11.0, 11.1, 11.2, 11.3, 11.4, 11.5, 11.6, 11.7, 11.8, 11.9. for which value of the internal resistance is the transferred power the highest? In general, for arbitrary values of the internal resistance and the emf, when is the power the highest? E2R ( 7 + R ) 2 EQUATION 5 R=11.0 4.0 x 11.0 PL = (11.4 + 11.0)2 PL =0.35 R=11.1 4.02 x 11.1 PL = (11.4 + 11.1)= PL = 0.35 R=11.2 4.0- x 11.2 PL = (11.4 + 11.2)= PL = 0.350849 R=11.3 4.02 x 11.3 P. = (11.4 + 11.3)= PL = 0.350870R=11.4 4.02 x 11.4 PL = (11.4 + 11.4)= PL = 0.350877 R=11.5 4.02 x 11.5 PL = (11.4 + 11.5)2 PL = 0.350870 R=11.6 4.0- x 11.6 PL = (11.4 + 11.6)2 PL = 0.350850 R=11.7 4.02 x 11.7 PL = (11.4 + 11.7)2 PL = 0.350818 R=11.8 4.02 x 11.8 P. = (11.4 + 11.8)= PL = 0.350668 R=11.9 4.03 x 11.9 PL = (11.4 + 11.9)= PL = 0.350715 All the P, values are relatively equivalent, however, they are slightly different our highest value is PL = 0.350877IS PL = 0.350877 The highest transferred power is at an internal resistance is R=11.4

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