(New tips are continually added to these pages. Check back in a few months' time for more)

Sample 3: TJC Prelim 2000 Q4 / GCE (S) Nov 99 Q4

The relation between current I and temperature T in a certain device is

I = AT²e^-B/T

where A and B are constants. Fig. 4.1 shows values of I and T obtained experimentally.

I/A	1.6x10^-5	1.7x10^-6	1.5x10^-7	1.9x10^-9	1.6x10^-10
T/K	740	700	660	600	570

(a) The value of B may be obtained from a linear graph of suitable functions of I and T. State the graph to be plotted and explain how the value of B is deduced from it.

(b) Obtain values of the required functions of I ant T. Plot the graph. Deduce the value of B from your graph.

Thinking Process:

Part (a)

Step 1:

First, recognise that there is an exponential function in the equation, hence you need to log_e both sides.

Step 2:

However, if you log_e both sides directly, you are going to end up with an equation involving log_eI, log_eT and B/T, which is not a linear graph, as required by the question.

Step 3:

Hence, before log_e both sides, divide throughout by T² so the equation becomes:

I/T² = Ae^-B/T

Step 4:

Now you can log_e both sides, to obtain:

log_e(I/T²) = log_eA - B/T

Step 5:

Plotting log_e(I/T²) against 1/T gives a linear graph with gradient -B and y-intercept log_eA

Part (b)

Step 1:

Re-tabulate the table adding three more rows for 1/T, I/T² and log_e(I/T²), as follows:

I/A	1.6x10^-5	1.7x10^-6	1.5x10^-7	1.9x10^-9	1.6x10^-10
T/K	740	700	660	600	570
1/T	1.35x10^-3	1.43x10^-3	1.52x10^-3	1.67x10^-3	1.75x10^-3
I/T²	2.9x10^-11	3.5x10^-12	3.4x10^-13	5.3x10^-15	4.9x10^-16
log_e(I/T²)	-24.3	-26.4	-28.7	-32.9	-35.2

Step 2:

Plot log_e (I/T²) against 1/T, as shown in the graph below:

Step 3:

From the graph, find the gradient, which is -27250. Thus -B = -27250. The value of B is 27250.

If need be, you can also find the value of log_eA, which is 12.5.

This gives the value of A as 2.68 x 10³.