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nameVoid
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According to Kirchoff's first law for electrical circuits V=RI+L(dI/dt) where constants V, R and L denote the electromotive force, the resistance, and the inductance, respectively, and I denotes the current at time t. If the electromotive force is terminated at time t=0 and if the current is I(0) at the instant of removal, prove that I=I(0)e^(-Rt/L)
[tex]
v=RI+L(\frac{dI}{dT})
[/tex]
[tex]
dt=\frac{LdI}{v-RI}
[/tex]
[tex]
t+b=Lln|v-RI|(-R)
[/tex]
[tex]
e^{\frac{t+b}{-LR}}=v-RI
[/tex]
[tex]
t-> v-> 0 , I=I(0)
[/tex]
[tex]
e^{\frac{b}{-LR}}=-RI(0)
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[tex]
-e^{\frac{t}{-LR}}(RI(0))=v-RI
[/tex]
[tex]
I=\frac{e^{\frac{t}{-LR}}RI(0)+v}{R}
[/tex]
im afraid to do anymore work on this but looks good to me so far
Homework Equations
The Attempt at a Solution
[tex]
v=RI+L(\frac{dI}{dT})
[/tex]
[tex]
dt=\frac{LdI}{v-RI}
[/tex]
[tex]
t+b=Lln|v-RI|(-R)
[/tex]
[tex]
e^{\frac{t+b}{-LR}}=v-RI
[/tex]
[tex]
t-> v-> 0 , I=I(0)
[/tex]
[tex]
e^{\frac{b}{-LR}}=-RI(0)
[/tex]
[tex]
-e^{\frac{t}{-LR}}(RI(0))=v-RI
[/tex]
[tex]
I=\frac{e^{\frac{t}{-LR}}RI(0)+v}{R}
[/tex]
im afraid to do anymore work on this but looks good to me so far
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