Given the expression I(R+a) = b, linearize it to plot a suitable graph, identify the variables for the axes, and determine the expressions for the gradient and y-intercept.

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Asked on April 10, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-04-10T21:27:37.593Z

You're on a roll — i) Plot a suitable graph Step 1: Linearize the given expression. The given expression is I(R+a) = b. Expand the expression: IR + Ia = b Rearrange the equation to match the linear form Y = mX + C. We want to plot R against (1)/(I). IR = b - Ia Divide by I: R = (b)/(I) - a This can be written as: R = b((1)/(I)) - a Comparing this to Y = mX + C: Y = R X = (1)/(I) m = b (gradient) C = -a (y-intercept) Step 2: Prepare a table of values for R and (1)/(I). Calculate the reciprocal of current, (1)/(I), for each given resistance R. | R/ | I/A | (1)/(I)/A^-1 | | :--------: | :----------: | :--------------------------: | | 10 | 0.200 | 5.00 | | 20 | 0.150 | 6.67 | | 30 | 0.120 | 8.33 | | 40 | 0.100 | 10.00 | | 50 | 0.086 | 11.63 | | 60 | 0.075 | 13.33 | | 80 | 0.060 | 16.67 | | 90 | 0.055 | 18.18 | | 100 | 0.050 | 20.00 | | 130 | 0.040 | 25.00 | Step 3: Describe the graph to be plotted. Plot a graph of R (y-axis) against (1)/(I) (x-axis). The graph should be a straight line. The gradient of this straight line will give the value of b. The y-intercept (the value of R when (1)/(I) = 0) will give the value of -a. ii) Obtain the values of a and b from the graph Step 1: Calculate the gradient (m) of the line. The gradient m = b. Using the first point (X_1, Y_1) = (5.00, 10) and the last point (X_2, Y_2) = (25.00, 130) from the table for the line of best fit: m = (Y_2 - Y_1)/(X_2 - X_1) = (130\, - 10\,)/(25.00\,A)^-1 - 5.00\,A^-1 m = (120\,)/(20.00\,A)^-1 = 6.0\, Therefore, b = 6.0\,V. Step 2: Calculate the y-intercept (C) of the line. The y-intercept C = -a. Using the equation Y = mX + C, which is R = b((1)/(I)) - a, and one point, for example (X, Y) = (5.00, 10): 10 = (6.0)(5.00) - a 10 = 30 - a a = 30 - 10 a = 20\, The values are: a = 20\, b = 6.0\,V iii) Physical significances of a and b The expression I(R+a) = b is analogous to the electromotive force (e.m.f.) equation for a simple circuit: I(R+r) = E, where E is the e.m.f. of the power source and r is its internal resistance. *

Accepted Answer · 2026-04-10T21:27:37.593Z

You're on a roll — i) Plot a suitable graph Step 1: Linearize the given expression. The given expression is I(R+a) = b. Expand the expression: IR + Ia = b Rearrange the equation to match the linear form Y = mX + C. We want to plot R against (1)/(I). IR = b - Ia Divide by I: R = (b)/(I) - a This can be written as: R = b((1)/(I)) - a Comparing this to Y = mX + C: Y = R X = (1)/(I) m = b (gradient) C = -a (y-intercept) Step 2: Prepare a table of values for R and (1)/(I). Calculate the reciprocal of current, (1)/(I), for each given resistance R. | R/ | I/A | (1)/(I)/A^-1 | | :--------: | :----------: | :--------------------------: | | 10 | 0.200 | 5.00 | | 20 | 0.150 | 6.67 | | 30 | 0.120 | 8.33 | | 40 | 0.100 | 10.00 | | 50 | 0.086 | 11.63 | | 60 | 0.075 | 13.33 | | 80 | 0.060 | 16.67 | | 90 | 0.055 | 18.18 | | 100 | 0.050 | 20.00 | | 130 | 0.040 | 25.00 | Step 3: Describe the graph to be plotted. Plot a graph of R (y-axis) against (1)/(I) (x-axis). The graph should be a straight line. The gradient of this straight line will give the value of b. The y-intercept (the value of R when (1)/(I) = 0) will give the value of -a. ii) Obtain the values of a and b from the graph Step 1: Calculate the gradient (m) of the line. The gradient m = b. Using the first point (X_1, Y_1) = (5.00, 10) and the last point (X_2, Y_2) = (25.00, 130) from the table for the line of best fit: m = (Y_2 - Y_1)/(X_2 - X_1) = (130\, - 10\,)/(25.00\,A)^-1 - 5.00\,A^-1 m = (120\,)/(20.00\,A)^-1 = 6.0\, Therefore, b = 6.0\,V. Step 2: Calculate the y-intercept (C) of the line. The y-intercept C = -a. Using the equation Y = mX + C, which is R = b((1)/(I)) - a, and one point, for example (X, Y) = (5.00, 10): 10 = (6.0)(5.00) - a 10 = 30 - a a = 30 - 10 a = 20\, The values are: a = 20\, b = 6.0\,V iii) Physical significances of a and b The expression I(R+a) = b is analogous to the electromotive force (e.m.f.) equation for a simple circuit: I(R+r) = E, where E is the e.m.f. of the power source and r is its internal resistance. *

Given the expression I(R+a) = b, linearize it to plot a suitable graph, identify the variables for the axes, and determine the expressions for the gradient and y-intercept.

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Given the expression I(R+a) = b, linearize it to plot a suitable graph, identify the variables for the axes, and determine the expressions for the gradient and y-intercept.

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