sid9221
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http://dl.dropbox.com/u/33103477/Resistor.png
For the first bit first I solved the equation
[tex]\frac{dI}{dt}=\frac{U-RI}{L}[/tex]
[tex]I(t)=\frac{U-C_2e^{\frac{-t}{L}}}{R}[/tex]
Now I put in T(0)=0 to work out the constant and I got
[tex]I(t)=\frac{U-Ue^{\frac{-t}{L}}}{R}[/tex]
Now here's the dodgy bit, I did not do physics in my final years at school so I no clue what an "Ohmic Resistor" is but here is my interpretation of the question
[tex]I(t)=\frac{75U}{100R}[/tex]
Working that out I got
[tex]t = -L log(0.25)[/tex]
Now for the final part:
My interpretation was that the maximum saturation level was U/R=200/50=4
So I worked out:
[tex]I(t)= 3.5[/tex]
Which gave me t=90.31
Does what I have done make sense cause I don't know much "higher level" physics so am just working from logic.
For the first bit first I solved the equation
[tex]\frac{dI}{dt}=\frac{U-RI}{L}[/tex]
[tex]I(t)=\frac{U-C_2e^{\frac{-t}{L}}}{R}[/tex]
Now I put in T(0)=0 to work out the constant and I got
[tex]I(t)=\frac{U-Ue^{\frac{-t}{L}}}{R}[/tex]
Now here's the dodgy bit, I did not do physics in my final years at school so I no clue what an "Ohmic Resistor" is but here is my interpretation of the question
[tex]I(t)=\frac{75U}{100R}[/tex]
Working that out I got
[tex]t = -L log(0.25)[/tex]
Now for the final part:
My interpretation was that the maximum saturation level was U/R=200/50=4
So I worked out:
[tex]I(t)= 3.5[/tex]
Which gave me t=90.31
Does what I have done make sense cause I don't know much "higher level" physics so am just working from logic.
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