Have you found an error or do you want to add more
information to these pages? You can contact me at the bottom of the home page. |
E = R | dq dt |
+ V_{C} |
E = CR | dV_{C} dt |
+ V_{C} |
1 CR |
dt = | 1 E - V_{C} |
dV_{C} | remember that | dt CR | = | 1 CR | dt |
∫ | 1 CR |
dt = | ∫ | 1 E - V_{C} |
dV_{C} |
∫ | 1 u |
dV_{C} |
∫ | 1 u |
-(du) | = - | ∫ | 1 u |
du = -ln(u) + K = -ln(E - V_{C}) + K (because u = E - V_{C}) |
∫ | 1 CR |
dt = | ∫ | 1 E - V_{C} |
dV_{C} |
t CR |
= -ln(E - V_{C}) + K (the general solution) |
0 CR |
= -ln(E - 0) + K |
t CR |
= -ln(E - V_{C}) + ln(E) = ln(E) - ln(E - V_{C}) = | ln | ( | E E - V_{C} |
) | because ln(a) - ln(b) = ln(a/b) |
t CR |
= ln | ( | E E - V_{C} |
) |
e^{t/CR} = | E E - V_{C} |
e^{-t/CR} = | E - V_{C} E |
E = iR + | q C |
E = R | dq dt |
+ | q C |
E - | q C |
= R | dq dt |
now I can divide both sides by R to get: |
E R |
- | q CR |
= | dq dt |
ECR - qR CR^{2} |
= | dq dt |
Now I can separate the variables: |
1 ECR - qR |
dq = | 1
CR^{2} |
dt This is an equation that I can integrate: |
∫ | 1 ECR - qR |
dq = | ∫ | 1
CR^{2} |
dt |
- | ln(ECR - qR) R |
= | t
CR^{2} |
+ K (where K is some constant) |
- | ln(ECR - 0R) R |
= |
0 CR^{2} |
+ K which simplifies to: |
- | ln(ECR) R |
= | K |
- | ln(ECR - qR) R |
= | t
CR^{2} |
- | ln(ECR) R |
but there are too many minus signs for my liking so I multiply everything by -1 |
ln(ECR - qR) R |
= - | t
CR^{2} |
+ | ln(ECR) R |
I notice that every fraction has at least one R term so I can multiply everything by R to make things a little simpler: |
ln(ECR - qR) = - | t
CR |
+ ln(ECR) I am going to put the logarithms together now: |
ln(ECR - qR) - ln(ECR) = - | t
CR |
ln | ( | ECR - qR ECR |
) | = - | t
CR |
then I raise e by the both sides of the equation to get rid of the natural logarithm. I can get at q after that: |
ECR - qR ECR |
= e^{-t/CR} | I notice that I can simplify the fraction: |
1 - | q EC |
= e^{-t/CR} |
i |
= | V_{R} R |
dq dt |
= |
E - V_{C} R |
= | E - qC^{-1} R |
Now I multiply by C / C (which is equal to 1) so I can get rid of C^{-1} |
dq dt |
= |
EC - q CR |
I transpose this to put the terms with q on one side |
1 EC - q |
dq = | 1 CR |
dt Now I place integral symbols in to it |
∫ | 1 EC - q |
dq = | ∫ | 1 CR |
dt |
-ln(EC - q) |
= - | t CR |
+ K |
-ln(EC - q) |
= - | t CR |
+ ln(E - C) |
1 - |
q EC |
= e^{-t/CR} |
Have you found an error or do you want to add more
information to these pages? You can contact me at the bottom of the home page. |