1. ## Log graph help

In a practical investigation of the discharging current for a capacitor, which was initially charged, the following values of current were obtained

Time
(seconds)
Current
(A)
0
1.00
0.1
0.47
0.2
0.23
0.3
0.10
0.4
0.05
0.5
0.02
0.6
0.01
The relationship between the charging current and time is believed to be of the form i=Ie *tiny*at

Reduce this to a linear relationship, plot a graph on the graph paper provided and from it find the values of I and a using both
a) 50 marks

Log10
b)Ln (loge)

Im at Bath university and im really struggling with log graphs. Any help would be very much apprieciated. this question is worth 30 marks and i cant get my head around it.

2. Originally Posted by sanchez
In a practical investigation of the discharging current for a capacitor, which was initially charged, the following values of current were obtained

Time
(seconds)
Current
(A)
0
1.00
0.1
0.47
0.2
0.23
0.3
0.10
0.4
0.05
0.5
0.02
0.6
0.01
The relationship between the charging current and time is believed to be of the form i=Ie *tiny*at
$i= Ie^{at}$

Reduce this to a linear relationship, plot a graph on the graph paper provided and from it find the values of I and a using both
a) 50 marks

Log10
b)Ln (loge)

Im at Bath university and im really struggling with log graphs. Any help would be very much apprieciated. this question is worth 30 marks and i cant get my head around it.
The problem says "plot a graph on the graph paper provided". I suspect you are provided with "semi-log graph paper" on which one of the axes actually shows "log y" rather than "y" so that rather than plotting " $y= Ae^{ax}$" you are plotting [tex]ln y= ln(Ae^{ax})= ax+ ln(A)). That appears, on the graph paper, as a straight line with "y- intercept" (0, ln(A)) and slope a. Since you can calculate both of those from the graph, you can get both A and a and so write out the equation $y= Ae^{ax}$.