% Get linear fit % Note that this is not perfect, since the fit does not go through the % origin. For a perfect fit, will need to use a more involved fit such as % lsqlin with the Matlab optimization toolbox % The first order poly fit will return the slope and intercept of the best % line fit [P]=polyfit(vm,im,1); % Construct curve for linear fit to plot v = linspace(-2.5,2.5,100); for i = 1:100 I(i) = v(i)*P(1); end R=1/P(1) plot(vm,im,'o',v,I,'LineWidth',1.0); xlabel('Voltage (V)','FontSize',14); ylabel('Current (mA)','FontSize',14); set(gca,'Fontsize',14,'LineWidth',1.0) 52. Utilize the fact that in the circuit of Fig. 2.39, the total power supplied by the voltage source must equal the total power absorbed by the two resistors to show that = (R2)/(R1 + VR2 VS R2). You may assume the same current flows through each element (a requirement of charge conservation). The power supplied by the voltage source is Vs I. The power absorbed by resistor is R1 I2R1. The power absorbed by resistor is R2 I2R2. Since we know that the total power supplied is equal to the total power absorbed, we may write: = + Vs I I2R1 I2R2 or = + Vs I R1 I R2 = + Vs I (R1 R2) By Ohm’s law, I = / R2 so that = Vs Solving for we find 53. For the circuit in Figure 2.39, suppose that the resistor R2 represents a very sensitive and expensive piece of electronics. To ensure that the equipment is not damaged, R1 is incorporated to represent a fuse, with a rating of 5A and resistance of 0.1 If the voltage source is 12 V, Ω.
% Get linear fit % Note that this is not perfect, since the fit does not go through the % origin. For a perfect fit, will need to use a more involved fit such as % lsqlin with the Matlab optimization toolbox % The first order poly fit will return the slope and intercept of the best % line fit [P]=polyfit(vm,im,1); % Construct curve for linear fit to plot v = linspace(-2.5,2.5,100); for i = 1:100 I(i) = v(i)*P(1); end R=1/P(1) plot(vm,im,'o',v,I,'LineWidth',1.0); xlabel('Voltage (V)','FontSize',14); ylabel('Current (mA)','FontSize',14); set(gca,'Fontsize',14,'LineWidth',1.0) 52. Utilize the fact that in the circuit of Fig. 2.39, the total power supplied by the voltage source must equal the total power absorbed by the two resistors to show that VR2 = VS (R2)/(R1 + R2). You may assume the same current flows through each element (a requirement of charge conservation). The power supplied by the voltage source is Vs I. The power absorbed by resistor R1 is I2R1. The power absorbed by resistor R2 is I2R2. Since we know that the total power supplied is equal to the total power absorbed, we may write: Vs I = I2R1 + I2R2 or Vs = I R1 + I R2 Vs = I (R1 + R2) By Ohm’s law, I = / R2 so that Vs = Solving for we find 53. For the circuit in Figure 2.39, suppose that the resistor R2 represents a very sensitive and expensive piece of electronics. To ensure that the equipment is not damaged, R1 is incorporated to represent a fuse, with a rating of 5A and resistance of 0.1 Ω. If the voltage source is 12 V,
Solution-Manual-for-Engineering-Circuit-Analysis-9th-Edition-By-Hayt.docx