# What Is The Gradient Of A Graph

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From the graph shown in Figure 6, it can be seen that the data has a linear relationship where y=0.0715 x. The extremely high r squared value of 0.9988 for the graph shows that a linear trendline is appropriate for the data set. The gradient of this graph can be used to calculate the constant of magnetism (k) which then can be compared against its actual value to determine the percentage error present in the results. As I_OUT MAX=(4π^2 fkN_1 N_2 I)/RL A, the gradient of a graph where peak output current is graphed against peak input current the gradient is equal to (4π^2 fkN_1 N_2)/RL A. Therfore, 0.0715=(4π^2 fkN_1 N_2 I)/RL A 0.0715=(4π^2*50*420*1000*k)/(11.9*0.109)*0.00056 (0.0715*1.9*0.109)/(4π^2*50*420*1000*0.00056)=k …show more content…

For example: 2 Volts RMS The percentage uncertainty in the peak induced EMF = 0.0025/0.066833*100 = 3.74% The percentage uncertainty in the peak input EMF = 0.0025/2.83*100 = 0.09% The percentage uncertainty in the radius of the solenoid = 0.00025/0.0135*100 = 1.85% The percentage uncertainty in the length of the solenoid = 0.00025/0.109*100 = 0.23% The percentage uncertainty in the combined resistance = 0.05/45.4*100 = 0.11% Therefore, total percentage uncertainty for input voltage of 2 Volts RMS= 6.02% % Uncertainty In 2 Volts 4 Volts 6 Volts 8 Volts 10 Volts 12 Volts Peak Induced EMF 3.74% 2.01% 1.41% 1.04% 0.82% 0.69% Peak Input EMF 0.09% 0.04% 0.03% 0.02% 0.02% 0.01% Radius of Solenoid 1.85% 1.85% 1.85% 1.85% 1.85% 1.85% Length of Solenoid 0.23% 0.23% 0.23% 0.23% 0.23% 0.23% Combined Resistance 0.11% 0.11% 0.11% 0.11% 0.11% 0.11% Total % Uncertainty 6.02% 4.24% 3.63% 3.25% 3.03% 2.90% Figure 8 – Percentage Uncertainty Calculations for Each Input Voltage 4.0 Discussion 4.1 Interpretation of Data Figure 5 on page 10 clearly demonstrates that the induced EMF in the secondary circuit increase as the voltage applied to the first solenoid increases. The graph produced for each voltage tested are sinusoidal functions, has a x-intercept (where the reading switches between positive and negative) every 0.01 seconds. Consequently, the time for one complete cycle (the period of the function) is 0.02 seconds. As frequency is equal to one divided by the