LR Circuit Analysis: Calculating Current and Gradient (0-0.6 seconds)

will.i.am2
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A coil connected across a d.c. supply will cause a spark to occur across a switch when turned off. The current flowing in the circuit at this moment is given by the equation i = IetR/L amps. Where I is the maximum circuit current, R the resistance of the coil and L the inductance of the coil.

(a) For a coil of resistance = 10Ω and Inductance 1.2 henrys through which the maximum current flowing was 5 amps, plot the graph for values of t between 0 and 0.6 seconds.
please refer to attached images.

(b)From the graph determine its gradient at t = 0.2 seconds.
26/0.2 = 130?
please refer to attached graph.

(c) Using calculus determine the gradient of the curve and from it the actual gradient when t = 0.2 seconds.
please refer to attached images.

So, any help would be appreciated.
thanks

****work is attached in form of images please don't remove it this time*****
 

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  • part c.png
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Remember when taking the derivative of an exponential, you just get an exponential with some constants out front. That might help in drawing your graph =)
 
Your derivative in part c is incorrect, as it was in the other problem you posted.

What is this derivative?
$$\frac{d}{dt}e^{tR/L}$$

BTW, it's reasonable to post images of tables and graphs, but you should post the work you did right here in the text pane, not as an image.
will.i.am2 said:
****work is attached in form of images please don't remove it this time*****
If you're referring to your post for the CR problem, the images are still there. No one removed them.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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