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RadiationX
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http://img360.imageshack.us/img360/2049/82009866sy5.th.jpg
To solve this circuit I'm going to use the differential equation approach. I'm concerned with the voltage across the capacitor at [tex]V_c(0^-)[/tex] and [tex]V_c(0^+)[/tex]
At position 1 before the switch is thrown [tex]V_c(0^-)=0[/tex]
At position 2 after switch is thrown I have the following from KCL:
[tex]\frac{6-V(t)}{12k} = C\frac{dv}{dt} + \frac{V(t)}{6k}[/tex]
This reduces to: [tex]\frac{dv}{dt} + 2.5V(t) = 5 [/tex]
My problem is that I know that this is correct but I don't know how to put it all together.
I know that the solution is of the form:
[tex]K_1 +K_2e^{-t/t_c}[/tex]
The answer is [tex]V(t)=1.33 -1.33e^{-2.5t}V[/tex]
How do I extract this from my work?
http://img360.imageshack.us/img360/2049/82009866sy5.th.jpg
To solve this circuit I'm going to use the differential equation approach. I'm concerned with the voltage across the capacitor at [tex]V_c(0^-)[/tex] and [tex]V_c(0^+)[/tex]
At position 1 before the switch is thrown [tex]V_c(0^-)=0[/tex]
At position 2 after switch is thrown I have the following from KCL:
[tex]\frac{6-V(t)}{12k} = C\frac{dv}{dt} + \frac{V(t)}{6k}[/tex]
This reduces to: [tex]\frac{dv}{dt} + 2.5V(t) = 5 [/tex]
My problem is that I know that this is correct but I don't know how to put it all together.
I know that the solution is of the form:
[tex]K_1 +K_2e^{-t/t_c}[/tex]
The answer is [tex]V(t)=1.33 -1.33e^{-2.5t}V[/tex]
How do I extract this from my work?
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